Question:
The unstable halide of the following is
The unstable halide of the following is
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For transition metal halides, smaller halides (like F\(^-\)) may destabilize lower oxidation states due to high lattice strain.
Updated On: Jun 20, 2026
- VI\(_2\)
- VBr\(_2\)
- VCl\(_2\)
- VF\(_2\)
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The Correct Option is D
Solution and Explanation
Step 1: Understand stability trend of metal halides.
Stability of metal halides depends on lattice energy and bond strength between metal and halogen. In transition metals, higher oxidation state halides are stabilized more by highly electronegative halogens like F. However, lower oxidation state fluorides can sometimes be unstable due to mismatch in size and bonding character.
Step 2: Analyze general trend of halides.
As we move from F to I: \[ F^- > Cl^- > Br^- > I^- \] Fluorides are generally more ionic and strongly bonded, but in some low oxidation states (like +2 for V), stability issues arise due to lattice strain and strong polarization effects.
Step 3: Compare given halides.
- VI\(_2\), VBr\(_2\), VCl\(_2\) are relatively more stable because larger halide ions stabilize lower oxidation state via better lattice compatibility.
- VF\(_2\) is comparatively less stable due to strong lattice distortion and high charge density mismatch between small F\(^-\) and V\(^{2+}\).
Step 4: Apply periodic stability reasoning.
For transition metal +2 halides, stability generally increases down the halogen group: \[ VI_2 > VBr_2 > VCl_2 > VF_2 \] Thus VF\(_2\) is the least stable (most unstable).
Step 5: Final comparison.
Since VF\(_2\) shows maximum instability among given options, it is the correct answer.
Final Answer: \[ \boxed{\text{VF}_2} \]
Stability of metal halides depends on lattice energy and bond strength between metal and halogen. In transition metals, higher oxidation state halides are stabilized more by highly electronegative halogens like F. However, lower oxidation state fluorides can sometimes be unstable due to mismatch in size and bonding character.
Step 2: Analyze general trend of halides.
As we move from F to I: \[ F^- > Cl^- > Br^- > I^- \] Fluorides are generally more ionic and strongly bonded, but in some low oxidation states (like +2 for V), stability issues arise due to lattice strain and strong polarization effects.
Step 3: Compare given halides.
- VI\(_2\), VBr\(_2\), VCl\(_2\) are relatively more stable because larger halide ions stabilize lower oxidation state via better lattice compatibility.
- VF\(_2\) is comparatively less stable due to strong lattice distortion and high charge density mismatch between small F\(^-\) and V\(^{2+}\).
Step 4: Apply periodic stability reasoning.
For transition metal +2 halides, stability generally increases down the halogen group: \[ VI_2 > VBr_2 > VCl_2 > VF_2 \] Thus VF\(_2\) is the least stable (most unstable).
Step 5: Final comparison.
Since VF\(_2\) shows maximum instability among given options, it is the correct answer.
Final Answer: \[ \boxed{\text{VF}_2} \]
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