Which of the following pair is not isoelectronic species?
(Atomic numbers Ho=67; Er=68; Yb=70; Lu=71; Eu=63; Tb=65; Tm=69)

Isoelectronic species possess the same number of electrons.

Number of electrons is calculated using:

$$\mathrm{Number\ of\ electrons = Atomic\ Number - Charge}$$

For:

$$\mathrm{Ho^{2+}}$$

$$\mathrm{67 - 2 = 65}$$

For:

$$\mathrm{Er^{3+}}$$

$$\mathrm{68 - 3 = 65}$$

Hence, they are isoelectronic.

For:

$$\mathrm{Yb^{2+}}$$

$$\mathrm{70 - 2 = 68}$$

For:

$$\mathrm{Lu^{3+}}$$

$$\mathrm{71 - 3 = 68}$$

Hence, they are isoelectronic.

For:

$$\mathrm{Eu^{2+}}$$

$$\mathrm{63 - 2 = 61}$$

For:

$$\mathrm{Tb^{4+}}$$

$$\mathrm{65 - 4 = 61}$$

Hence, they are isoelectronic.

For:

$$\mathrm{Tb^{2+}}$$

$$\mathrm{65 - 2 = 63}$$

For:

$$\mathrm{Tm^{4+}}$$

$$\mathrm{69 - 4 = 65}$$

Hence, they are not isoelectronic.

Therefore, the pair which is not isoelectronic is:

$$\mathrm{Tb^{2+}\ and\ Tm^{4+}}$$

Correct option:

$$\mathrm{D}$$