Question:
The tetra atomic molecules/ions with different shapes are
The tetra atomic molecules/ions with different shapes are
Show Hint
According to VSEPR theory:
\[
AX_4 \rightarrow \text{tetrahedral}
\]
and
\[
AX_4E \rightarrow \text{see-saw}
\]
where \(E\) represents a lone pair.
Updated On: Jun 24, 2026
- \(TeCl_4,\ SeF_4\)
- \(CH_4,\ PCl_4^+\)
- \(SF_4,\ NH_4^+\)
- \(SiH_4,\ CCl_4\)
Show Solution
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The Correct Option is C
Solution and Explanation
Step 1: Determine the shape of \(SF_4\).
In \[ SF_4, \] sulphur has \[ 6 \] valence electrons.
It forms four bonds with fluorine atoms and has one lone pair. Thus, the steric number is \[ 5 \] According to VSEPR theory, the electron pair geometry is trigonal bipyramidal and the molecular shape becomes see-saw
Step 2: Determine the shape of \(NH_4^+\).
In \[ NH_4^+, \] nitrogen forms four bonds and has no lone pair.
Hence, the geometry is tetrahedral
Step 3: Compare the shapes in other options.
\[ TeCl_4 \] and \[ SeF_4 \] both have see-saw shape.
\[ CH_4 \] and \[ PCl_4^+ \] both are tetrahedral.
\[ SiH_4 \] and \[ CCl_4 \] are also tetrahedral.
Only \[ SF_4 \] and \[ NH_4^+ \] have different shapes.
Step 4: Final conclusion.
Therefore, the correct answer is \[ \boxed{SF_4,\ NH_4^+} \]
In \[ SF_4, \] sulphur has \[ 6 \] valence electrons.
It forms four bonds with fluorine atoms and has one lone pair. Thus, the steric number is \[ 5 \] According to VSEPR theory, the electron pair geometry is trigonal bipyramidal and the molecular shape becomes see-saw
Step 2: Determine the shape of \(NH_4^+\).
In \[ NH_4^+, \] nitrogen forms four bonds and has no lone pair.
Hence, the geometry is tetrahedral
Step 3: Compare the shapes in other options.
\[ TeCl_4 \] and \[ SeF_4 \] both have see-saw shape.
\[ CH_4 \] and \[ PCl_4^+ \] both are tetrahedral.
\[ SiH_4 \] and \[ CCl_4 \] are also tetrahedral.
Only \[ SF_4 \] and \[ NH_4^+ \] have different shapes.
Step 4: Final conclusion.
Therefore, the correct answer is \[ \boxed{SF_4,\ NH_4^+} \]
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