Question

Which of the following statement(s) is/are correct?
$S_1:Se{F}_6$ has the octahedral structure and $P{F}_5$ has $s{p}^3{d}^2$ hybridization.
$S_2$: The percentage of s-character in the orbital forming S−S bonds and P−P bonds in $S_8$ and $P_4$ molecules respectively are 25% and 33.3%.
$S_3:$ $\left[Co\right(N{H}_3{)}_6{]}^{3+}$ is an inner orbital complex as well as diamagnetic in behaviour.

• A. Only $S_1$
• B. $S_2$ and $S_3$
• C. $S_1$ and $S_3$
• D. Only $S_3$

Answer 1

The correct option is D Only $S_3$

$S_1:$ The central Se atom has 6 bond pairs of electrons and 0 lone pairs of electrons. It

undergoes $s{p}^3{d}^2$ hybridisation, which results in octahedral geometry. In case of $P{F}_5$, the

central P atom has 5 sets of electron pairs (all of which are bonding) and $s{p}^3{d}^2$ hybridization

corresponds to 6 sets of electron pairs.Therefore $P{F}_5$ is $s{p}^{3}d$ hybridized. Hence, statement 1 is

false.

$S_2:S_8\left(s{p}^3{d}^2\right)$ and $P_4\left(s{p}^3\right)$ both has $25\%$ s- character.

$S_3:$ Orbitals in $C{o}^{+3}$ ion

$\uparrow \downarrow \uparrow \uparrow \uparrow \uparrow$

$3d$ $4s$ $4p$

Making Hybridized orbitals

$\uparrow \downarrow \uparrow \downarrow \uparrow \downarrow$

$d^{2}s{p}^3$

$\left[Co\right(N{H}_3{)}_6{]}^{3+}$

$\uparrow \downarrow \uparrow \downarrow \uparrow \downarrow$

$\uparrow \downarrow \uparrow \downarrow \uparrow \downarrow \uparrow \downarrow \uparrow \downarrow \uparrow \downarrow$

$6$ pairs of electrons from $6N{H}_3$ molecules.

Inner '$d$' orbitals are used in hybridization so its inner orbital complex. There are no

unpaired electrons so that it is diamagnetic.

Option D is correct.