Question

Identify the correct statement.
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• A. The central atoms in both \(CH_4\) and \(SF_4\) are in a state of \(sp^3\) hybridisation
• B. The resultant dipole moment of \(NF_3\) is greater than that of \(NH_3\)
• C. The \(C-O\) bond length in \(CO_2\) molecule is \(110 \, pm\) because of the inductive effect \((+I)\) of oxygen
• D. A molecule of the type \(AB_5E\), where the central atom has 5 bond pairs and 1 lone pair, has square pyramidal geometry

Hint:

In VSEPR theory, \(AB_5E\) has 6 electron domains, so its electron-pair geometry is octahedral and molecular shape is square pyramidal.

Answer 1

The Correct Option is D

Step 1: Analyse statement (A).
In \(CH_4\), carbon has four bond pairs and no lone pair.
Therefore, carbon is \(sp^3\) hybridised and the shape is tetrahedral.
However, in \(SF_4\), sulphur has four bond pairs and one lone pair.
So, the steric number is:
\[ 4 + 1 = 5 \] Thus, sulphur in \(SF_4\) is \(sp^3d\) hybridised, not \(sp^3\).
Therefore, statement (A) is incorrect.

Step 2: Analyse statement (B).
Both \(NH_3\) and \(NF_3\) have pyramidal geometry.
In \(NH_3\), the bond dipoles and lone pair dipole act in nearly the same direction, so the resultant dipole moment is comparatively high.
In \(NF_3\), the \(N-F\) bond dipoles act opposite to the lone pair dipole, so the resultant dipole moment is reduced.
Hence, dipole moment of \(NH_3\) is greater than that of \(NF_3\).
Therefore, statement (B) is incorrect.

Step 3: Analyse statement (C).
In \(CO_2\), carbon forms two double bonds with oxygen atoms.
The structure is:
\[ O=C=O \] The \(C-O\) bond length in \(CO_2\) is short due to double bond character, not due to \(+I\) effect of oxygen.
Oxygen is highly electronegative and shows \(-I\) effect, not \(+I\) effect.
Therefore, statement (C) is incorrect.

Step 4: Analyse statement (D).
For a molecule of type \(AB_5E\), the central atom has 5 bond pairs and 1 lone pair.
Total electron pairs around the central atom are:
\[ 5 + 1 = 6 \] So, the electron pair geometry is octahedral.

Step 5: Determine the molecular geometry of \(AB_5E\).
In an octahedral arrangement, if one position is occupied by a lone pair, the remaining five bond pairs form a square pyramidal shape.
Thus, \(AB_5E\) type molecule has square pyramidal geometry.

Step 6: Final selection.
Only statement (D) is correct because \(AB_5E\) has octahedral electron-pair geometry and square pyramidal molecular geometry.
Final Answer:
The correct statement is:
\[ \boxed{\text{(D) A molecule of the type } AB_5E \text{ has square pyramidal geometry}} \]