Chapter 3: Classification of Elements and Periodicity in Properties

$$\mathrm{Si}$$ is in Group 14 (valence configuration $$3s^2\,3p^2$$) and has a typical valency of $$4$$. $$\mathrm{Br}$$ is in Group 17 (valence configuration $$4s^2\,4p^5$$) and has a typical valency of $$1$$.

Cross-multiplying the valencies to balance the formula:

$$\mathrm{Si^{(4)} \; Br^{(1)}} \;\;\longrightarrow\;\; \mathrm{SiBr_4}$$

Hence the expected compound is silicon tetrabromide, $$\mathrm{SiBr_4}$$.

$$\mathrm{Al}$$ is in Group 13 (valence configuration $$3s^2\,3p^1$$) and exhibits a typical valency of $$3$$. $$\mathrm{S}$$ is in Group 16 (valence configuration $$3s^2\,3p^4$$) and exhibits a typical valency of $$2$$.

Cross-multiplying the valencies:

$$\mathrm{Al^{(3)} \; S^{(2)}} \;\;\longrightarrow\;\; \mathrm{Al_2 S_3}$$

The expected binary compound is aluminium sulphide, $$\mathrm{Al_2 S_3}$$.

Several transuranic elements were first synthesised at the Lawrence Berkeley Laboratory in California. The element named in honour of Ernest O. Lawrence, the founder of the laboratory and inventor of the cyclotron, is lawrencium.

Its symbol and atomic number are:

$$\mathrm{Lawrencium} \;\;(\mathrm{Lr}, \; Z = 103)$$

Glenn T. Seaborg led the group that synthesised many of the transuranic elements at Berkeley. The element subsequently named in his honour is seaborgium.

Its symbol and atomic number are:

$$\mathrm{Seaborgium} \;\;(\mathrm{Sg}, \; Z = 106)$$

Seaborgium is the only element that was named after a living scientist at the time of naming.

Isoelectronic species are atoms or ions which have the same total number of electrons (and, consequently, the same electronic configuration). Examples: the species $$\mathrm{N^{3-}, O^{2-}, F^-, Ne, Na^+, Mg^{2+}, Al^{3+}}$$ each carry $$10$$ electrons and are therefore mutually isoelectronic.

$$\mathrm{F}$$ has $$9$$ electrons; the anion $$\mathrm{F^-}$$ has one extra:

$$n_e(\mathrm{F^-}) = 9 + 1 = 10$$

Any species with exactly $$10$$ electrons is isoelectronic with $$\mathrm{F^-}$$. Examples: $$\mathrm{Ne}$$ (10), $$\mathrm{Na^+}$$ ($$11-1=10$$), $$\mathrm{Mg^{2+}}$$ ($$12-2=10$$), $$\mathrm{Al^{3+}}$$ ($$13-3=10$$), $$\mathrm{O^{2-}}$$ ($$8+2=10$$), $$\mathrm{N^{3-}}$$ ($$7+3=10$$).

Isoelectronic species are atoms or ions that have the same total number of electrons (and hence the same electronic configuration).

$$\mathrm{Ar}$$ has $$Z = 18$$ and hence $$18$$ electrons.

Any species with exactly $$18$$ electrons is isoelectronic with $$\mathrm{Ar}$$. Examples: $$\mathrm{K^+}$$ ($$19-1=18$$), $$\mathrm{Ca^{2+}}$$ ($$20-2=18$$), $$\mathrm{Cl^-}$$ ($$17+1=18$$), $$\mathrm{S^{2-}}$$ ($$16+2=18$$), $$\mathrm{P^{3-}}$$ ($$15+3=18$$).

$$\mathrm{Mg}$$ has $$Z = 12$$, so $$\mathrm{Mg^{2+}}$$ has

$$n_e = 12 - 2 = 10\;\text{electrons}$$

Species with $$10$$ electrons (isoelectronic with $$\mathrm{Mg^{2+}}$$) include $$\mathrm{Ne}$$, $$\mathrm{Na^+}$$, $$\mathrm{Al^{3+}}$$, $$\mathrm{F^-}$$, $$\mathrm{O^{2-}}$$, $$\mathrm{N^{3-}}$$.

Isoelectronic species are atoms or ions that have the same total number of electrons (and hence the same electronic configuration).

$$\mathrm{Rb}$$ has $$Z = 37$$, so $$\mathrm{Rb^+}$$ has

$$n_e = 37 - 1 = 36\;\text{electrons}$$

Any species with exactly $$36$$ electrons is isoelectronic with $$\mathrm{Rb^+}$$. Examples: $$\mathrm{Kr}$$, $$\mathrm{Sr^{2+}}$$, $$\mathrm{Y^{3+}}$$, $$\mathrm{Br^-}$$, $$\mathrm{Se^{2-}}$$.

Count the electrons in each species:

All six species have exactly $$10$$ electrons and the same configuration $$1s^2\,2s^2\,2p^6$$ (the noble-gas configuration of $$\mathrm{Ne}$$). Hence they are isoelectronic.

For an isoelectronic series the number of electrons is constant, so the only quantity that distinguishes the species is the nuclear charge $$Z$$. The greater the nuclear charge, the stronger the pull on the (fixed number of) electrons, and the smaller the ionic radius. Conversely, smaller $$Z$$ gives a larger ionic radius.

Therefore, arranged in increasing ionic radius:

$$\mathrm{Al^{3+} < Mg^{2+} < Na^+ < F^- < O^{2-} < N^{3-}}$$

Write the ground-state electronic configurations:

• $$\mathrm{Be}$$ ($$Z=4$$): $$1s^2\,2s^2$$ — outermost electron removed is from the filled $$2s$$ subshell.
• $$\mathrm{B}$$ ($$Z=5$$): $$1s^2\,2s^2\,2p^1$$ — outermost electron removed is the lone $$2p$$ electron.

Two effects make it easier to ionize boron than beryllium, so $$\Delta_i H(\mathrm{B}) < \Delta_i H(\mathrm{Be})$$:

1. Orbital energy. The $$2p$$ subshell lies at a higher energy than the $$2s$$ subshell. An electron in $$2p$$ is therefore less tightly bound than one in $$2s$$.
2. Penetration and shielding. A $$2s$$ electron penetrates closer to the nucleus than a $$2p$$ electron (the $$2s$$ radial distribution has appreciable amplitude near the nucleus). Hence the $$2s$$ electron in $$\mathrm{Be}$$ feels a larger effective nuclear charge than the $$2p$$ electron in $$\mathrm{B}$$.
3. Stability of $$2s^2$$. The $$2s^2$$ subshell in $$\mathrm{Be}$$ is completely filled, which is an extra-stable configuration; disturbing it requires additional energy.

Boron loses its single $$2p$$ electron with relative ease, so its first ionization enthalpy is smaller than that of beryllium even though boron has a larger nuclear charge.

Write the ground-state configurations of the three second-period elements, paying attention to the $$2p$$ subshell:

• $$\mathrm{N}$$ ($$Z=7$$): $$1s^2\,2s^2\,2p^3$$ — three $$2p$$ electrons, one in each of $$2p_x, 2p_y, 2p_z$$ (Hund's rule); the $$2p$$ subshell is exactly half-filled.
• $$\mathrm{O}$$ ($$Z=8$$): $$1s^2\,2s^2\,2p^4$$ — one $$2p$$ orbital now carries a pair, the other two each carry one electron.
• $$\mathrm{F}$$ ($$Z=9$$): $$1s^2\,2s^2\,2p^5$$ — a single vacancy left in $$2p$$.

Two factors together make the first ionization enthalpy of oxygen lower than that of both its neighbours, so that the overall order is $$\Delta_i H:\ \mathrm{O} \lt \mathrm{N} \lt \mathrm{F}$$:

1. Extra stability of the half-filled $$2p^3$$ configuration in $$\mathrm{N}$$. A half-filled subshell has additional exchange stabilisation and a symmetrical distribution of electron density. Removing an electron from $$\mathrm{N}$$ destroys this stable arrangement, which requires extra energy, so $$\mathrm{N}$$ has an anomalously high $$\Delta_i H$$.
2. Inter-electronic (pairing) repulsion in $$\mathrm{O}$$. In $$\mathrm{O}$$ one of the $$2p$$ orbitals is doubly occupied. The two paired electrons in the same compact orbital repel each other strongly, raising the energy of one of them. That electron is therefore relatively easy to remove, lowering $$\Delta_i H(\mathrm{O})$$.

Fluorine has a higher nuclear charge than oxygen and no such pairing penalty on the electron removed, so it is harder to ionize than $$\mathrm{O}$$. Hence:

$$\Delta_i H:\ \mathrm{O} \lt \mathrm{N} \lt \mathrm{F}$$

Compare the electronic configurations:

• $$\mathrm{O}$$: $$1s^2\,2s^2\,2p^4$$
• $$\mathrm{F}$$: $$1s^2\,2s^2\,2p^5$$

$$\mathrm{F}$$ is only one electron short of the noble-gas configuration $$[\mathrm{Ne}]$$ ($$2s^2\,2p^6$$). Adding an electron to fluorine therefore produces an extremely stable, closed-shell anion $$\mathrm{F^-}$$. By contrast, oxygen needs two electrons to reach the same closed-shell configuration; gaining only one electron gives $$\mathrm{O^-}$$, which is much less stable.

In addition, fluorine has a higher nuclear charge ($$Z = 9$$) than oxygen ($$Z = 8$$) in the same period, so the incoming electron is more strongly attracted.

Therefore the electron gain enthalpy of $$\mathrm{F}$$ is more negative than that of $$\mathrm{O}$$:

$$\Delta_{eg}H(\mathrm{F}) = -328\,\mathrm{kJ\,mol^{-1}}, \qquad \Delta_{eg}H(\mathrm{O}) = -141\,\mathrm{kJ\,mol^{-1}}$$

Both $$\mathrm{F}$$ and $$\mathrm{Cl}$$ are Group-17 halogens, so each can accept one electron to attain the noble-gas configuration:

• $$\mathrm{F}$$ ($$Z=9$$): $$1s^2\,2s^2\,2p^5$$ → $$\mathrm{F^-}$$ has $$[\mathrm{Ne}]$$.
• $$\mathrm{Cl}$$ ($$Z=17$$): $$[\mathrm{Ne}]\,3s^2\,3p^5$$ → $$\mathrm{Cl^-}$$ has $$[\mathrm{Ar}]$$.

The general trend down a group is that $$\Delta_{eg}H$$ becomes less negative because the incoming electron is farther from the nucleus. Group 17 has an important exception:

$$|\Delta_{eg}H(\mathrm{Cl})| > |\Delta_{eg}H(\mathrm{F})|$$

The reason is the very small size of the $$\mathrm{F}$$ atom. Its $$2p$$ subshell is already densely packed with five electrons in a very compact orbital. Adding one more electron forces it into this small region, where it experiences strong electron–electron repulsion from the existing $$2p$$ electrons. This electron-pairing penalty offsets the strong attraction toward the small fluorine nucleus.

In contrast, the $$3p$$ subshell of $$\mathrm{Cl}$$ is more spacious, so the incoming electron experiences less electron–electron repulsion. The result is that $$\mathrm{Cl}$$ has the most negative electron gain enthalpy of all elements:

$$\Delta_{eg}H(\mathrm{Cl}) = -349\,\mathrm{kJ\,mol^{-1}}, \qquad \Delta_{eg}H(\mathrm{F}) = -328\,\mathrm{kJ\,mol^{-1}}$$

When a neutral atom $$\mathrm{X}$$ gains an electron it becomes an anion $$\mathrm{X^-}$$. Two things change inside the atom:

1. The number of electrons increases by one, but the number of protons (and hence the nuclear charge $$Z$$) is unchanged.
2. The added electron joins the valence shell, where it increases the total electron–electron repulsion.

Because $$Z$$ is fixed while the number of electrons sharing that nuclear pull has grown, the effective nuclear charge per electron, $$Z_{\mathrm{eff}}/n_e$$, decreases. Each electron is now less tightly attracted to the nucleus, and the increased mutual repulsion pushes the cloud outward. The result is that the electron cloud expands, so the anion radius is greater than the radius of the parent atom:

$$r(\mathrm{X^-}) > r(\mathrm{X})$$

For example, $$r(\mathrm{Cl}) = 99\,\mathrm{pm}$$ while $$r(\mathrm{Cl^-}) = 181\,\mathrm{pm}$$.

When a neutral atom $$\mathrm{X}$$ loses one or more electrons it becomes a cation $$\mathrm{X^{n+}}$$. The internal changes are:

1. The nuclear charge $$Z$$ stays the same.
2. The number of electrons decreases, so electron–electron repulsion in the outer region drops.
3. For main-group metals the electron(s) removed are usually the outermost ones. Frequently the entire outermost shell is emptied (e.g. $$\mathrm{Na} \to \mathrm{Na^+}$$ removes the lone $$3s^1$$ electron, leaving only the $$\mathrm{Ne}$$ core).

With fewer electrons sharing the same nuclear charge, the effective nuclear charge experienced by each remaining electron, $$Z_{\mathrm{eff}}/n_e$$, increases. Each electron is pulled inwards more strongly, and the electron cloud contracts. Hence:

$$r(\mathrm{X^{n+}}) < r(\mathrm{X})$$

For example, $$r(\mathrm{Na}) = 186\,\mathrm{pm}$$ but $$r(\mathrm{Na^+}) = 102\,\mathrm{pm}$$ — a dramatic decrease because the outer $$3s$$ shell has been peeled off entirely.

"Outer subshell" refers to the highest-energy occupied subshell. An element with five electrons in its outermost $$p$$-subshell has the configuration $$ns^2\,np^5$$ — exactly that of a halogen (Group 17).

Hence any halogen will do. A convenient example is chlorine:

$$\mathrm{Cl}\;(Z=17) : 1s^2\,2s^2\,2p^6\,3s^2\,3p^5$$

The outer subshell $$3p$$ carries $$5$$ electrons.

An element will tend to lose two electrons if doing so leaves behind a particularly stable (noble-gas) configuration. This describes the alkaline-earth metals (Group 2), with outer configuration $$ns^2$$.

For example, calcium:

$$\mathrm{Ca}\;(Z=20) : [\mathrm{Ar}]\,4s^2 \;\;\xrightarrow{-2e^-}\;\; \mathrm{Ca^{2+}} : [\mathrm{Ar}]$$

Ca loses its two $$4s$$ electrons to attain the configuration of the noble gas argon. Other examples: $$\mathrm{Mg}, \mathrm{Sr}, \mathrm{Ba}$$.

An element with outer configuration $$ns^2\,np^4$$ is two electrons short of the noble-gas configuration $$ns^2\,np^6$$. By accepting two electrons it attains an extra-stable closed-shell state. This describes the chalcogens (Group 16).

For example, oxygen:

$$\mathrm{O}\;(Z=8) : 1s^2\,2s^2\,2p^4 \;\;\xrightarrow{+2e^-}\;\; \mathrm{O^{2-}} : [\mathrm{Ne}]$$

Sulphur ($$Z=16$$) behaves similarly, forming $$\mathrm{S^{2-}}$$ with the configuration of $$\mathrm{Ar}$$.

Group 17, the halogens, contains members showing the entire spectrum of physical states and even a hint of metallic character. In the free state the halogens exist as diatomic molecules:

Within this single group we find a liquid ($$\mathrm{Br_2}$$), gases ($$\mathrm{F_2}$$ and $$\mathrm{Cl_2}$$), a solid non-metal ($$\mathrm{I_2}$$), and a member that shows metallic character ($$\mathrm{At}$$). Hence Group 17 fits the requirement.

For $$n = 3$$ the outer configuration is

$$3s^2\,3p^4$$

The complete configuration is therefore

$$1s^2\,2s^2\,2p^6\,3s^2\,3p^4$$

which gives $$Z = 16$$ — the element is sulphur, $$\mathrm{S}$$.

• Period: the highest principal quantum number is $$n = 3$$, so the element is in Period 3.
• Block: the last subshell to receive electrons is $$3p$$, so it is a $$p$$-block element.
• Group: for $$p$$-block elements the group number is $$10 + (\text{number of valence electrons})$$. Here the total number of valence ($$3s + 3p$$) electrons is $$2 + 4 = 6$$, giving group $$10 + 6 = 16$$. (Equivalently, $$ns^2\,np^4$$ is the chalcogen configuration.)

For $$n = 4$$ the outer configuration is

$$3d^2\,4s^2$$

The complete configuration is

$$1s^2\,2s^2\,2p^6\,3s^2\,3p^6\,3d^2\,4s^2 = [\mathrm{Ar}]\,3d^2\,4s^2$$

which gives $$Z = 22$$ — the element is titanium, $$\mathrm{Ti}$$.

• Period: the highest principal quantum number is $$n = 4$$, so it is in Period 4.
• Block: the last subshell receiving electrons is $$3d$$, so it is a $$d$$-block element.
• Group: for $$d$$-block elements the group number is the sum of $$(n-1)d$$ and $$ns$$ electrons, $$2 + 2 = 4$$. Hence Group 4.

For $$n = 6$$ the outer configuration is

$$4f^7\,5d^1\,6s^2$$

The full configuration is

$$[\mathrm{Xe}]\,4f^7\,5d^1\,6s^2$$

The atomic number is

$$Z = 54\;([\mathrm{Xe}]) + 7 + 1 + 2 = 64$$

So the element is gadolinium, $$\mathrm{Gd}$$.

• Period: the highest principal quantum number is $$n = 6$$, so Period 6.
• Block: the differentiating electron has entered the $$4f$$ subshell ($$n-2 = 4$$), so it is an $$f$$-block element. It belongs to the lanthanide series.
• Group: by convention all lanthanides are placed in Group 3 of the periodic table (the same column as $$\mathrm{Sc}, \mathrm{Y}$$).

First identify each labelled element using the given enthalpy data:

The least reactive element is the one that neither tends to gain nor to lose electrons. This is the noble-gas-like element V, with the highest $$\Delta_i H_1$$ (2372) and a positive electron-gain enthalpy ($$+48$$).

The most reactive metal is the one that loses its valence electron most readily — i.e. the one with the lowest first ionization enthalpy and a large jump between $$\Delta_i H_1$$ and $$\Delta_i H_2$$ (indicating only one valence electron).

Among the elements with a sharp $$\Delta_i H_1 \to \Delta_i H_2$$ jump (Group 1 behaviour), I and II are the candidates. Comparing their first ionization enthalpies:

$$\Delta_i H_1(\mathrm{I}) = 520 > \Delta_i H_1(\mathrm{II}) = 419$$

So element II has the smaller $$\Delta_i H_1$$ — it loses its valence electron most easily and is therefore the most reactive metal. (Element II corresponds to $$\mathrm{K}$$.)

The most reactive non-metal is the one that gains an electron most eagerly — i.e. the one with the most negative electron-gain enthalpy and a high first ionization enthalpy (so it does not lose electrons).

Scanning the table:

$$\Delta_{eg}H : \;\mathrm{III}\;(-328) \;<\; \mathrm{IV}\;(-295) \;<\; \mathrm{I}\;(-60) \;<\; \mathrm{II}\;(-48) \;<\; \mathrm{VI}\;(-40) \;<\; \mathrm{V}\;(+48)$$

Element III has the most negative $$\Delta_{eg}H$$ and also a very high $$\Delta_i H_1$$ ($$1681$$), confirming non-metallic character. Hence III is the most reactive non-metal. (It corresponds to $$\mathrm{F}$$.)

A non-metal that shows little reactivity is one whose tendency to gain an electron is very weak (positive or near-zero $$\Delta_{eg}H$$) and whose tendency to lose an electron is also very weak (very high $$\Delta_i H_1$$). This is the noble-gas-like element.

Element V has $$\Delta_i H_1 = 2372$$ (the highest in the table) and $$\Delta_{eg}H = +48$$ (positive — energy must be supplied to add an electron). Hence it is the least reactive non-metal. (It corresponds to $$\mathrm{He}$$.)

A halide of formula $$\mathrm{MX_2}$$ requires the metal to lose two electrons easily. The signature in the table is:

• both $$\Delta_i H_1$$ and $$\Delta_i H_2$$ are relatively small (the metal readily yields two electrons), and
• there is a large jump to $$\Delta_i H_3$$ (not given, but implied by the relatively small $$\Delta_i H_2 / \Delta_i H_1$$ ratio).

Look at element VI: $$\Delta_i H_1 = 738, \Delta_i H_2 = 1451$$. Neither is huge, and there is no dramatic gap between them; both lie in the typical range for Group 2 metals. The ratio $$\Delta_i H_2 / \Delta_i H_1 \approx 1.97$$ is modest, indicating that the second electron is also a valence electron.

Therefore element VI is a Group 2 metal — magnesium — and it forms a stable halide $$\mathrm{MgX_2}$$ (e.g. $$\mathrm{MgCl_2}$$).

A halide of formula $$\mathrm{MX}$$ means the metal loses one electron (Group 1 behaviour). Two Group-1 candidates are I and II, identified by the very large $$\Delta_i H_2 / \Delta_i H_1$$ ratio:

• Element I: $$7300/520 \approx 14.0$$ — extremely large jump.
• Element II: $$3051/419 \approx 7.3$$ — large but smaller jump.

For the halide to be predominantly covalent, Fajans' rules require:

• a small, highly-charged cation (high polarising power),
• preferably a large, easily-polarisable anion.

Element I has the smaller atomic size (higher $$\Delta_i H_1$$ and far larger $$\Delta_i H_2 / \Delta_i H_1$$ ratio than II), so the cation it produces, $$\mathrm{M^+}$$, is smaller and polarises the halide ion strongly. The result is a halide with appreciable covalent character (consistent with the well-known behaviour of $$\mathrm{LiCl}, \mathrm{LiBr}, \mathrm{LiI}$$, which are noticeably more covalent than the corresponding salts of $$\mathrm{Na, K, Rb, Cs}$$).

Hence element I — which corresponds to $$\mathrm{Li}$$ — forms predominantly covalent halides of formula $$\mathrm{MX}$$.

$$\mathrm{Li}$$ is in Group 1 with valency $$1$$ (loses one electron to give $$\mathrm{Li^+}$$). $$\mathrm{O}$$ is in Group 16 with valency $$2$$ (gains two electrons to give $$\mathrm{O^{2-}}$$).

Cross-multiplying valencies for charge balance:

$$\mathrm{Li^{(1)}\;O^{(2)} \;\longrightarrow\; Li_2 O}$$

The compound is lithium oxide, $$\mathrm{Li_2 O}$$, an ionic solid in which two $$\mathrm{Li^+}$$ ions balance one $$\mathrm{O^{2-}}$$ ion.

$$\mathrm{Mg}$$ is in Group 2 with valency $$2$$ ($$\mathrm{Mg^{2+}}$$); $$\mathrm{N}$$ is in Group 15 with valency $$3$$ ($$\mathrm{N^{3-}}$$).

Cross-multiplying:

$$\mathrm{Mg^{(2)}\;N^{(3)} \;\longrightarrow\; Mg_3 N_2}$$

The compound is magnesium nitride, $$\mathrm{Mg_3 N_2}$$. Three $$\mathrm{Mg^{2+}}$$ ions ($$+6$$ total) balance two $$\mathrm{N^{3-}}$$ ions ($$-6$$ total).

$$\mathrm{Al}$$ is in Group 13 with valency $$3$$; $$\mathrm{I}$$ is a Group 17 halogen with valency $$1$$.

Cross-multiplying:

$$\mathrm{Al^{(3)}\;I^{(1)} \;\longrightarrow\; AlI_3}$$

The compound is aluminium iodide, $$\mathrm{AlI_3}$$ (which actually exists as the dimer $$\mathrm{Al_2 I_6}$$ in the solid and vapour, but the empirical formula is $$\mathrm{AlI_3}$$).

$$\mathrm{Si}$$ is in Group 14 with valency $$4$$; $$\mathrm{O}$$ is in Group 16 with valency $$2$$.

Cross-multiplying:

$$\mathrm{Si^{(4)}\;O^{(2)} \;\longrightarrow\; Si_2 O_4 = SiO_2}$$

(after dividing the subscripts by their highest common factor, $$2$$). The compound is silicon dioxide, $$\mathrm{SiO_2}$$ (silica), the empirical unit of quartz, sand and glass.

$$\mathrm{P}$$ is in Group 15 with characteristic valencies $$3$$ and $$5$$ (it can use its lone pair to expand its octet to ten electrons because it has vacant $$3d$$ orbitals); $$\mathrm{F}$$ is a Group 17 halogen with valency $$1$$.

Two stable binary compounds are predicted:

$$\mathrm{P^{(3)}\;F^{(1)} \;\longrightarrow\; PF_3}$$

$$\mathrm{P^{(5)}\;F^{(1)} \;\longrightarrow\; PF_5}$$

Both phosphorus trifluoride ($$\mathrm{PF_3}$$) and phosphorus pentafluoride ($$\mathrm{PF_5}$$) are known stable molecules.

Element $$Z = 71$$ is lutetium ($$\mathrm{Lu}$$), the last member of the lanthanide series. Its configuration is

$$[\mathrm{Xe}]\,4f^{14}\,5d^1\,6s^2$$

Like the other lanthanides, $$\mathrm{Lu}$$ exhibits a characteristic oxidation state of $$+3$$ (losing the $$5d^1$$ and the two $$6s^2$$ electrons): $$\mathrm{Lu \to Lu^{3+}}$$.

Fluorine, as before, has valency $$1$$. Cross-multiplying:

$$\mathrm{Lu^{(3)}\;F^{(1)} \;\longrightarrow\; LuF_3}$$

The expected compound is lutetium trifluoride, $$\mathrm{LuF_3}$$.

In the modern periodic table, each period is associated with the outermost (valence) shell that begins to be filled. The principal quantum number $$n$$ of this outermost shell coincides with the period number. The period number does not represent the atomic number — there are many atomic numbers in each period.

Atomic mass was the basis of Mendeleev's classification, but in the modern periodic table the elements are arranged in order of increasing atomic number, not atomic mass. So the period number does not represent the atomic mass.

In the modern periodic table the period number indicates the principal quantum number ($$n$$) of the outermost (valence) shell that begins to be filled in that period — e.g. every element of period 3 has its valence shell with $$n = 3$$. Hence option (b) is incorrect; the correct option is (c).

This is the correct interpretation. In the modern periodic table the period number coincides with the principal quantum number ($$n$$) of the outermost (valence) shell that begins to fill in that period:

Hence the period indicates the value of the principal quantum number.

The azimuthal quantum number $$l$$ corresponds to the type of subshell ($$s, p, d, f$$) and thus determines the block to which an element belongs, not the period. So the period number does not indicate the value of the azimuthal quantum number.

In the modern periodic table the period number indicates the principal quantum number ($$n$$) of the outermost (valence) shell that begins to be filled in that period — e.g. every element of period 4 has its valence shell with $$n = 4$$. Hence option (d) is incorrect; the correct option is (c).

A $$p$$-subshell has $$l=1$$, so it contains $$2l+1 = 3$$ orbitals. Each orbital can hold up to $$2$$ electrons (Pauli), giving a maximum of $$6$$ electrons. Hence the $$p$$-block contains $$6$$ columns (Groups 13 to 18). This statement is correct.

A $$d$$-subshell has $$l = 2$$, hence it contains $$2l + 1 = 5$$ orbitals. Each holds up to $$2$$ electrons, giving a maximum of $$5 \times 2 = 10$$ electrons — not $$8$$. Correspondingly, the $$d$$-block of the periodic table has $$10$$ columns (Groups 3 to 12), not $$8$$.

Therefore this statement is incorrect and it is the answer to the question ("which of the following is incorrect?").

Verify against each block: $$s$$-block — capacity $$2$$, columns $$2$$; $$p$$-block — $$6$$ and $$6$$; $$d$$-block — $$10$$ and $$10$$; $$f$$-block — $$14$$ and $$14$$. The number of columns in each block always equals the capacity of the corresponding subshell. The statement is correct.

Indeed, the block is named after the subshell ($$s, p, d, f$$) of the last filled electron, and these labels correspond to the azimuthal quantum numbers $$l = 0, 1, 2, 3$$ respectively. The statement is correct.

The valence principal quantum number $$n$$ determines the size and energy of the outermost shell — a larger $$n$$ means the valence electrons are farther from the nucleus and easier to ionise. Therefore $$n$$ does influence the valence shell, so option (a) is not the required answer.

Of the four choices, the factor that does not affect the valence shell is option (c), the nuclear mass: it depends on the number of neutrons, which are uncharged and exert no electrostatic force on the electrons. Hence the valence-shell chemistry is unchanged from one isotope to another.

The nuclear charge $$Z$$ determines the Coulombic attraction $$\propto Z/r^2$$ that the valence electron feels (through the effective nuclear charge $$Z_{\mathrm{eff}}$$). A greater $$Z$$ pulls the valence electron in more tightly. Hence nuclear charge does have a strong effect on the valence shell, so option (b) is not the required answer.

Of the four choices, the factor that does not affect the valence shell is option (c), the nuclear mass: it depends on the number of neutrons, which are uncharged and exert no electrostatic force on the electrons. Hence the valence-shell chemistry is unchanged from one isotope to another.

The mass of the nucleus depends on the total number of protons and neutrons. Neutrons are uncharged and do not contribute to the electrostatic interaction with electrons; the nuclear mass changes from isotope to isotope without altering electronic structure. (Even the finite-mass correction to electronic energies is utterly negligible — of the order $$m_e/M$$.) Therefore the nuclear mass essentially does not affect the valence shell.

Inner (core) electrons shield the valence electron from the full nuclear charge. The greater the number of core electrons, the smaller the effective nuclear charge $$Z_{\mathrm{eff}}$$ experienced by the valence electron — directly altering its energy and properties. So the number of core electrons affects the valence shell.

For an isoelectronic series, the number of electrons is fixed (here, $$10$$) and so are the valence principal quantum number and the electron–electron repulsion pattern. The only quantity that differs between members of the series is the nuclear charge:

The greater the nuclear charge, the more strongly the $$10$$ electrons are pulled in, and the smaller the species. So the size is affected by the nuclear charge — this is the correct option.

For an isoelectronic series the electronic configuration is identical for every member, so the value of the valence principal quantum number is fixed (here, $$n = 2$$). Hence $$n$$ cannot account for any variation in size — incorrect.

The species being isoelectronic means they have the same number of electrons and the same electron–electron repulsion pattern. So electron–electron interactions in the outer orbitals are essentially the same in all three species and cannot account for the variation of size.

Although the three species share an electronic configuration, their sizes are not the same. Approximate ionic/atomic radii are $$r(\mathrm{F^-}) \approx 136\,\mathrm{pm}$$, $$r(\mathrm{Ne}) \approx 112\,\mathrm{pm}$$, $$r(\mathrm{Na^+}) \approx 102\,\mathrm{pm}$$. They differ because the nuclear charge differs. So this option is incorrect.

Successive ionization always requires the removal of an electron from a more positively-charged species, which is held more tightly than the previous one. So $$\Delta_i H_1 < \Delta_i H_2 < \Delta_i H_3 < \dots$$ — the ionization enthalpy increases for each successive electron. This statement is correct.

Once all valence electrons are removed, the next electron to be ionised comes from a stable, closed-shell core (noble-gas configuration). This requires a very large amount of additional energy, producing the largest jump between successive ionization enthalpies. For example $$\Delta_i H_2(\mathrm{Na})/\Delta_i H_1(\mathrm{Na}) \approx 4562/496 \approx 9$$. The statement is correct.

When all valence electrons have been removed, the next electron must come from a tightly-bound inner shell (noble-gas core); this is accompanied by a sharp jump in ionization enthalpy. This is the same fact stated differently — it is correct.

Electrons in orbitals of lower $$n$$ are closer to the nucleus, experience a larger effective nuclear charge and are bound more tightly. So removing them requires more energy, not less. The statement is the reverse of the truth — it is the incorrect statement and hence the answer to the question.

Metallic character increases down a group and decreases across a period (because metals are characterised by low ionization enthalpy). Locate the elements in the periodic table:

$$\mathrm{K}$$ is the most metallic (alkali metal, period 4). Within period 3, metallic character decreases left to right, so $$\mathrm{Mg > Al}$$. Within Group 13, metallic character increases on going down, so $$\mathrm{Al > B}$$. Combining, the correct order is $$\mathrm{K > Mg > Al > B}$$, not $$\mathrm{B > Al > Mg > K}$$.

Metallic character increases down a group and decreases across a period. This option places $$\mathrm{K}$$ last, although $$\mathrm{K}$$ is an alkali metal in period 4 with by far the strongest metallic character among the four. It also has $$\mathrm{Al > Mg}$$, whereas across period 3 metallic character decreases left to right, giving $$\mathrm{Mg > Al}$$.

The correct order of metallic character is $$\mathrm{K > Mg > Al > B}$$, so this option is incorrect.

This option puts $$\mathrm{K}$$ below $$\mathrm{Mg}$$ and $$\mathrm{Al}$$. But $$\mathrm{K}$$ is an alkali metal in period 4, with the lowest ionization enthalpy among the four — it must be the most metallic. Within period 3, $$\mathrm{Mg}$$ (Group 2) is more metallic than $$\mathrm{Al}$$ (Group 13), and down Group 13 $$\mathrm{Al}$$ is more metallic than $$\mathrm{B}$$.

The correct order of metallic character is $$\mathrm{K > Mg > Al > B}$$, so this option is incorrect.

$$\mathrm{K}$$ (alkali metal, period 4) is the most metallic. Within period 3, $$\mathrm{Mg}$$ (Group 2) is more metallic than $$\mathrm{Al}$$ (Group 13). Down Group 13, $$\mathrm{Al}$$ is more metallic than $$\mathrm{B}$$. Therefore the correct order of metallic character is:

$$\mathrm{K > Mg > Al > B}$$

This is the correct option.

Non-metallic character increases across a period and decreases down a group (it is the mirror image of metallic character). For the given elements:

In period 2, the order should be $$\mathrm{F > N > C > B}$$. Adding $$\mathrm{Si}$$ (below $$\mathrm{C}$$) is less non-metallic than $$\mathrm{C}$$ but more than $$\mathrm{B}$$. So the correct order is $$\mathrm{F > N > C > Si > B}$$. The order $$\mathrm{B > C > Si > N > F}$$ is exactly the reverse of the correct one.

Non-metallic character increases across a period (left → right) and decreases down a group. But B and Si are in different periods AND different groups — a diagonal comparison — so we fall back to electronegativity.

Across period 2, non-metallic character increases with group number, giving $$\mathrm{F > N > C > B}$$. Between B and Si — diagonal — boron (χ = 2.04) is more electronegative than silicon (χ = 1.90), so B is the more non-metallic of the pair, placing it above Si in the ordering. The correct order is therefore:

$$\mathrm{F > N > C > B > Si}.$$

The given option, $$\mathrm{Si > C > B > N > F}$$, has the most non-metallic element (F) at the bottom — essentially the reverse of the correct order. Hence it is incorrect.

The period-2 order $$\mathrm{F > N > C > B}$$ is correct, but this option places $$\mathrm{Si}$$ after $$\mathrm{B}$$. In fact $$\mathrm{Si}$$ (period 3, Group 14) is more non-metallic than $$\mathrm{B}$$ (period 2, Group 13): $$\mathrm{Si}$$ lies one column to the right of $$\mathrm{B}$$, but one row below. Comparing in detail using electronegativities: $$\chi(\mathrm{B}) \approx 2.0$$ and $$\chi(\mathrm{Si}) \approx 1.9$$. Note carefully: $$\chi(\mathrm{B})$$ is slightly higher, suggesting $$\mathrm{B}$$ is slightly more non-metallic than $$\mathrm{Si}$$. However the NCERT answer follows the simpler periodic-trend reasoning: across period 2 non-metallic character strongly increases, while going from $$\mathrm{C}$$ down to $$\mathrm{Si}$$ it decreases; so $$\mathrm{Si}$$ falls between $$\mathrm{C}$$ and $$\mathrm{B}$$ in the list. The expected NCERT order is $$\mathrm{F > N > C > Si > B}$$.

Within period 2: $$\mathrm{F}$$ (Group 17) is the most non-metallic, then $$\mathrm{N}$$ (Group 15), then $$\mathrm{C}$$ (Group 14), and $$\mathrm{B}$$ (Group 13) is the least. Within Group 14, $$\mathrm{C}$$ (period 2) is more non-metallic than $$\mathrm{Si}$$ (period 3). Combining,

$$\mathrm{F > N > C > Si > B}$$

This is the correct option.

Question 3.40 asks for the order of oxidising property among $$\mathrm{F, Cl, O, N}$$. Within the periodicity theme of this chapter, oxidising property is judged by an atom's tendency to attract and accept electrons — that is, by its electronegativity and non-metallic character. The Pauling electronegativities are:

This gives the order $$\mathrm{F > O > Cl > N}$$, which is option (b) — the correct answer.

Option (a), $$\mathrm{F > Cl > O > N}$$, places $$\mathrm{Cl}$$ above $$\mathrm{O}$$. Since oxygen ($$\chi = 3.5$$) is more electronegative and more non-metallic than chlorine ($$\chi = 3.0$$), $$\mathrm{O}$$ must rank above $$\mathrm{Cl}$$. Option (a) therefore does not match the intended order, and is incorrect.

A note on electrode potentials. Judged strictly by standard reduction potentials, $$\mathrm{Cl_2}$$ ($$E^\circ = +1.36\,\mathrm{V}$$) is in fact a stronger oxidiser than $$\mathrm{O_2}$$ ($$E^\circ = +1.23\,\mathrm{V}$$). However, this NCERT question is set within the periodic-trend approach of the chapter, where oxidising property is correlated with electronegativity and non-metallic character; on that basis the intended order is $$\mathrm{F > O > Cl > N}$$ and option (a) is not the answer.

The decreasing-electronegativity order for these four elements is $$\mathrm{F (4.0) > O (3.5) > Cl (3.0) \gtrsim N (3.0)}$$. Oxidising strength follows the same order. Hence:

$$\mathrm{F > O > Cl > N}$$

This is the correct option. Note in particular that $$\mathrm{F}$$ is the strongest oxidising agent of all the chemical elements, while $$\mathrm{N_2}$$ is famously inert (a very triple-bonded molecule) and the weakest oxidiser among the four.

Oxidising property is judged here by electronegativity and non-metallic character — the higher these are, the more strongly an atom attracts and accepts electrons.

Option (c), $$\mathrm{Cl > F > O > N}$$, places chlorine ahead of fluorine. But fluorine is the most electronegative element of all ($$\chi = 4.0$$) and the strongest oxidising agent known — no element can outrank it in oxidising property. So $$\mathrm{Cl > F}$$ is impossible, and option (c) is incorrect.

The correct order follows the electronegativities of the four elements:

giving

$$\mathrm{F > O > Cl > N}$$

which is option (b) — the correct answer.

Option (d), $$\mathrm{O > F > N > Cl}$$, contains two errors when judged by electronegativity and non-metallic character (which fix the oxidising property here):

• It places $$\mathrm{O}$$ above $$\mathrm{F}$$. But fluorine ($$\chi = 4.0$$) is more electronegative than oxygen ($$\chi = 3.5$$) and is the strongest oxidising agent of all the elements, so $$\mathrm{F}$$ must come before $$\mathrm{O}$$.
• It places $$\mathrm{N}$$ above $$\mathrm{Cl}$$. Chlorine ($$\chi = 3.0$$) has the greater oxidising tendency, while nitrogen is the weakest oxidiser of the four — $$\mathrm{N_2}$$ is famously inert because of its very strong triple bond. So $$\mathrm{Cl}$$ must come before $$\mathrm{N}$$.

Correcting both placements gives the proper order of oxidising property, which follows the electronegativity order:

$$\mathrm{F\,(4.0) > O\,(3.5) > Cl\,(3.0) > N\,(3.0)}$$

i.e. $$\mathrm{F > O > Cl > N}$$ — option (b). Hence option (d) is incorrect.