JEE Chemistry 2026 Question with Solution

Given: Two statements are given about bond lengths and bond order.

Find: Which option correctly identifies the truth values of Statement I and Statement II.

Statement I: Analyze the species $\mathrm{BF_4^-}$, $\mathrm{SiF_4}$, $\mathrm{XeF_4}$ and $\mathrm{SF_4}$.

• $\mathrm{BF_4^-}$ is tetrahedral, so all $\mathrm{B-F}$ bonds are equivalent.
• $\mathrm{SiF_4}$ is tetrahedral, so all $\mathrm{Si-F}$ bonds are equivalent.
• $\mathrm{XeF_4}$ is square planar, so all $\mathrm{Xe-F}$ bonds are equivalent.
• $\mathrm{SF_4}$ has seesaw geometry, so axial and equatorial $\mathrm{S-F}$ bonds are different.

Therefore, only $\mathrm{SF_4}$ has unequal $\mathrm{E-F}$ bond lengths. The number is $1$, not $3$.

So, Statement I is false.

Statement II: Use molecular orbital bond order.

Given values:

• $\mathrm{O_2^-}$ has bond order $1.5$
• $\mathrm{O_2^{2-}}$ has bond order $1$
• $\mathrm{F_2}$ has bond order $1$
• $\mathrm{O_2^+}$ has bond order $2.5$

Thus, $\mathrm{O_2^+}$ has the highest bond order.

So, Statement II is true.

Therefore, the correct option is B: Statement I is false but Statement II is true.

Given: Statement I depends on molecular geometry and symmetry, while Statement II depends on molecular orbital bond order.

Find: Evaluate both statements separately.

For Statement I, unequal bond lengths appear when the central atom experiences different bonding environments such as axial and equatorial positions.

• In $\mathrm{BF_4^-}$, the geometry is tetrahedral and all fluorine atoms are equivalent.
• In $\mathrm{SiF_4}$, the geometry is tetrahedral and all fluorine atoms are equivalent.
• In $\mathrm{XeF_4}$, the geometry is square planar with two lone pairs opposite each other, so all four $\mathrm{Xe-F}$ bonds remain equivalent.
• In $\mathrm{SF_4}$, the geometry is seesaw, derived from trigonal bipyramidal electron arrangement with one lone pair. Axial and equatorial bonds are not equivalent.

Hence only one species has unequal bond lengths, so Statement I is false.

For Statement II, compare bond orders:

The largest value is for $\mathrm{O_2^+}$, so Statement II is true.

Therefore, the correct option is B.