Question

Consider the value of all four quantum numbers for the last electron and spin multiplicity $(2s+1)$,
for the two given elements ${}^{\prime }{X}^{\prime }$ and ${}^{\prime }{Y}^{\prime }$ in their ground state:

$\begin{array}{cccccc}& n& l& m& s& [2s+1]\\ X:& 2& 0& 0& +\frac{1}{2}& 1\\ Y:& 2& 1& -1& -\frac{1}{2}& 4\end{array}$

Then according to given information the correct statement is:

• A. The bond angle $(H-Y-H)$ of the possible hydride of element $Y$ is less than ${109}^{\circ }{28}^{{}^{\prime }}$
• B. The possible halide of ${}^{\prime }{X}^{\prime }$ has two vacant $p-$ orbitals on its central atom.
• C. Magnetic moment of $Y$ is greater than $X$
• D. Elements $X$ and $Y$ exhibit a single oxidation state.

Answer 1

Answer: (A), (B), (C)

Magnetic moment of $Y$ is greater than $X$

For $X:n=2,l=0$
$\therefore$ Valence subshell is $2s$.
$\because 2s+1=1$
$\therefore s=0$. This is possible if two electrons with different spins are present.
$\therefore$ Outer electronic configuration is $2s^2$
$\therefore X$ is $Be$
For $Yn=2,l=1$
$\therefore$ Valence subshell is $2p$
$\because 2s+1=4$
$\therefore s=\frac{3}{2}$. This is possible when three electrons with same spin are present in a subshell.
$\therefore$ Outer electonic configuration is $2s^{2}2p^3$.
$\therefore Y$ is $N$.

a) Possible hydride of $N$ is $N{H}_3$. In $N{H}_3$, $H-N-H$ bond angle is less than ${109}^{\circ }{28}^{{}^{\prime }}$ because of strong lone pair-bond pair repulsion.

b) In $Be{X}_2(X:\text{halogen})$, hybridisation of $Be$ is $sp$ so two $p$ orbitals are unhybridised (of $Be$ atom).

c) Magnetic moment $=\sqrt{n(n+2)}B.M.$, where $n=$ no. of unpaired electrons. Value of $n$ for $Be$ is $O$ and for $N$ it is equal to 3. So magnetic moment of $Be=\sqrt{0(0+2)=0}$
Magnetic moment of $N=\sqrt{3(3+2)}=\sqrt{15}B.M.$

d) $N$ can exhibit variable oxidation states.
Eg. $\to$ In $N{F}_3$, it is +3 in $N{H}_3$ it is -3.

So, options a, b and c are correct.