Which one of the following metals forms interstitial hydride easily?

We begin by recalling that transition metals, especially those belonging to the earlier part of the d-block, can absorb hydrogen atoms into the voids (interstices) of their metallic lattices, thereby giving what are called interstitial hydrides. In such hydrides, the small $$H$$ atoms occupy the tetrahedral or octahedral holes of the metal lattice without markedly disturbing the close-packed metal framework.

The general representation of an interstitial hydride is written as

$$M + \dfrac{x}{2}\,H_2 \;\longrightarrow\; MH_x,$$

where $$M$$ stands for the metal and $$x$$ is a non-stoichiometric, often fractional, value because the exact number of hydrogen atoms accommodated depends on temperature, pressure and the radius of the interstices.

For the formation of such hydrides two main factors are crucial:

1. The metal lattice must possess sufficiently large holes so that the radius ratio $$\dfrac{r_H}{r_M}$$ is small enough (about 0.4 or less) to let the hydrogen atoms slip in.
2. The metal should have a relatively low electronegativity so that the bonding is predominantly metallic with a partial $$M \, \delta^{+}$$ and $$H \, \delta^{-}$$ character.

Metals situated toward the left of the first transition series (Sc, Ti, V, Cr, etc.) meet these requirements better than the later members (Fe, Co, Ni, Cu, Zn) because their atomic radii are larger and their lattices provide more spacious interstitial sites.

Among the four given choices—$$\text{Cr}$$, $$\text{Fe}$$, $$\text{Mn}$$ and $$\text{Co}$$—chromium is the earliest element in the series, lying in Group 6 with a comparatively large atomic radius (128 pm) and a bcc lattice that can readily take up hydrogen. Text-book data show that chromium forms non-stoichiometric hydrides approximately of composition $$\text{CrH}_{0.8}$$ under moderate conditions, whereas manganese, iron and cobalt require much more severe pressures and still absorb less hydrogen.

So, when ease of hydride formation is compared, chromium stands out:

$$\text{Cr} + x\,H_2 \;\longrightarrow\; \text{CrH}_x \quad\text{(occurs most readily)}.$$

Hence, the correct answer is Option A.