Question

Select the correct statement(s) for the above given structures.

• A. In $S_4{N}_4{H}_4$, the hydrogens never go on sulphurs while in $S_4{N}_4{F}_4$, fluorines get preferably attached with sulphur atoms.
• B. The highly electronegative fluorines on being placed on sulphur atoms can appreciably contract the 3d-orbital to facilitate $N\left(2p_n\right)\to S\left(3d_n\right)\pi$-bonding.
• C. If in $S_4{N}_4{H}_4$, hydrogens go to the sulphurs, no $\pi$-bonding occurs through the 3d-orbitals of $S$.
• D. $N-H$ bond is stronger than the $S-H$ bonds.

Answer 1

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

The correct options are
A The highly electronegative fluorines on being placed on sulphur atoms can appreciably contract the 3d-orbital to facilitate $N\left(2p_n\right)\to S\left(3d_n\right)\pi$-bonding.
B In $S_4{N}_4{H}_4$, the hydrogens never go on sulphurs while in $S_4{N}_4{F}_4$, fluorines get preferably attached with sulphur atoms.
C If in $S_4{N}_4{H}_4$, hydrogens go to the sulphurs, no $\pi$-bonding occurs through the 3d-orbitals of $S$.
D $N-H$ bond is stronger than the $S-H$ bonds.

All the four statements are correct.
(A) In $S_4{N}_4{H}_4$, the hydrogens never go on sulphurs, they are on $N$ atoms while in $S_4{N}_4{F}_4$, fluorines get preferably attached with sulphur atoms, they do not go with $N$ atoms. Both $N$ and $F$ are highly electronegative and do not form stable bonds easily.
(B) The highly electronegative fluorines on being placed on sulphur atoms can appreciably contract the 3d-orbital of facilitate $N\left(2p_n\right)\to S\left(3d_n\right)\pi$-bonding.
(C) If in $S_4{N}_4{H}_4$ hydrogens go to the sulphurs, no effective $\pi$-bonding occurs through the 3d-orbitals of $S$.
(D) $N-H$ bond is stronger than the $S-H$ bonds because of high electronegativity difference.