Question

Match the following:

Column - I		Column - II
A	$s{p}^3$	i	$CoN{H}_{36}{}^{3+}$
B	$ds{p}^2$	ii	$NiC{O}_4$
C	$s{p}^3{d}^2$	iii	$PtN{H}_{32}C{l}_2$
D	$d^{2}s{p}^3$	iv	$Co{F}_6^{3-}$

• A. A- i ,B- ii ,C- iv ,D- iii
• B. A- i ,B-iii ,C- iv ,D-ii
• C. A- ii ,B-iii ,C- i ,D-iv
• D. A- ii ,B-iii ,C- iv,D- i

Answer 1

Answer: (D)

I. ${\left[Co{\left(N{H}_3\right)}_6\right]}^{3+}$
Here $Co$ is present as $C{o}^{3+}$ ion. $N{H}_3$ being strong field ligand pair up the unpaired
electrons of $Co.$
${}_{27}Co={\mathrm{Ar}}^7{d}^{7}4s^2$
$\therefore C{o}^{3+}=A{r}^3{d}^{6}4s^0$

II. $NiC{O}_4$
Here $Ni$ is present in its ground state. $CO$ being a strong field ligand, pair up the unpaired electrons of $Ni$
${}_{28}Ni=\text{Ar}3d^{8}4s^2$

III. $PtN{H}_{32}C{l}_2$
Here $Pt$ is present as $P{t}^{2+}$ ion.
${}_{78}Pt=Xe5d^{9}6s^1$
$P{t}^{2+}=Xe5d^{8}6s^0$

IV. $Co{F}_6^{3-}$
Here $Co$ is present as $C{o}^{3+}$ ion, $F$ being weak field ligand is unable to pair up its unpaired electrons.
$C{o}^{3+}=[Ar]3d^{6}4s^0$