The hybridizations of atomic orbitals of nitrogen in \text{NO}_2^ + ,\text{NO}_3^ - and \text{NH}_4^ + respectively are

  • Option 1)

    sp, sp3 and sp2

  • Option 2)

    sp2, sp3 and sp

  • Option 3)

    sp, sp2 and sp3

  • Option 4)

    sp2, sp and sp3

 

Answers (1)

None

Dettermination of shape of molecules using VSEPR Theory -

calculate \:X

X=(No.\;of\:valence\:electrons\:of\:central\:atom\:)+(No.\:of\:other\:atom\:)+(Negative\:charge\:on\:the\:molecule)-(Positive\:charge\:on\:the\:molecule)

-

 

 X=\frac{1}{2}(VE+MA-C+a)

For NONO_2^{+}, X=\frac{1}{2}(5+0-1)=2\: i.e\: sp\: hybridised

For NO_{3}^{-}, X=\frac{1}{2}(5+0+1)=3 i.e\: sp^{2}\: hybridisation

For NH_{4}^{+}, X=\frac{1}{2}(5+4-1)=4\: i.e\: sp^{3}\: hybridisation


Option 1)

sp, sp3 and sp2

This option is incorrect

Option 2)

sp2, sp3 and sp

This option is incorrect

Option 3)

sp, sp2 and sp3

This option is correct

Option 4)

sp2, sp and sp3

This option is incorrect

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