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In the molecules $\mathrm{CH}_4, \mathrm{NF}_3, \mathrm{NH}_4^{+}$and $\mathrm{H}_2 \mathrm{O}$
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The correct answer is:
all have same hybridisation of centre of atom
all have same hybridisation of centre of atom
$\mathrm{CH}_4 ; \mathrm{bp}=4 ; s p^3$-hybridisation
$\mathrm{NF}_3 ; \mathrm{bp}=3 ; l p=1 ; s p^3$-hybridisation
$\mathrm{NH}_4^{+} ; \mathrm{bp}=4 ; l p=0 ; s p^3$-hybridisation
$\mathrm{H}_2 \mathrm{O} ; \mathrm{bp}=2 ; l p=2 ; s p^3$-hybridisation
Hence, all have the same slope.
$\mathrm{NF}_3 ; \mathrm{bp}=3 ; l p=1 ; s p^3$-hybridisation
$\mathrm{NH}_4^{+} ; \mathrm{bp}=4 ; l p=0 ; s p^3$-hybridisation
$\mathrm{H}_2 \mathrm{O} ; \mathrm{bp}=2 ; l p=2 ; s p^3$-hybridisation
Hence, all have the same slope.
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