Bond order of NO molecule is

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4.Chemical Bonding and Molecular Structure

medium

The bond order of $NO$ molecule is

$1$

$2$

$2.5$

$3$

Solution

(c) ${\rm{B}}{\rm{.O}}{\rm{.}} = \frac{{{\rm{No}}{\rm{. of bonding }}{e^ – } – {\rm{ No}}{\rm{. of antibonding }}{e^ – }}}{2}$ $ = \frac{{8 – 3}}{2} = \frac{5}{2} = 2.5$.

Standard 11

Chemistry

Bond order is a concept in the molecular orbital theory. It depends on the number of electrons in the bonding and antibonding orbitals. Which of the following statements is true about it ? The bond order

easy

(AIIMS-1980)

Match List $I$ with List $II$ :

List $I$ (Molecule Species)

List $I$ ( Property Shape )

$A$ $\mathrm{SO}_2 \mathrm{Cl}_2$ $I$ Paramagnetic

$B$ $NO$ $II$ Diamagnetic

$C$ $\mathrm{NO}_2^{-}$ $III$ Tetrahedral

$D$ $\mathrm{I}_3^{-}$ $IV$ Linear

Choose the correct answer from the options given below :

medium

(JEE MAIN-2024)

Match each of the diatomic molecules in Column $I$ with its property / properties in Column $II$.

Column $I$ Column $II$

$(A)$ $\mathrm{B}_2$ $(p)$ Paramagnetic

$(B)$ $\mathrm{N}_2$ $(q)$ Undergoes oxidation

$(C)$ $\mathrm{O}_2^{-}$ $(r)$ Undergoes reduction

$(D)$ $\mathrm{O}_2$ $(s)$ Bond order $\geq 2$

$(t)$ Mixing of ' $\mathrm{s}$ ' and ' $\mathrm{p}$ ' orbitals

hard

(IIT-2009)

In which pair, or pairs, is the stronger bond found in the first species

$(a)$ $O_2^ – ,{O_2}$ $(b)$ ${N_2},N_2^ +$ $(c)$ $NO^+, NO^-$

medium

Match $List-I$ with $List-II$.

$List-I$ $List-II$

$(A)$ $\Psi_{ MO }=\Psi_{ A }-\Psi_{ B }$ $(I)$ Dipole moment

$(B)$ $\mu=Q \times I$ $(II)$ Bonding molecular orbital

$(C)$ $\frac{N_{b}-N_{a}}{2}$ $(III)$ Anti-bonding molecualr orbital

$(D)$ $\Psi_{ MO }=\Psi_{ A }+\Psi_{ B }$ $(IV)$ Bond order

medium

(JEE MAIN-2022)

List $I$ (Molecule Species)	List $I$ ( Property Shape )
$A$ $\mathrm{SO}_2 \mathrm{Cl}_2$	$I$ Paramagnetic
$B$ $NO$	$II$ Diamagnetic
$C$ $\mathrm{NO}_2^{-}$	$III$ Tetrahedral
$D$ $\mathrm{I}_3^{-}$	$IV$ Linear

Column $I$	Column $II$
$(A)$ $\mathrm{B}_2$	$(p)$ Paramagnetic
$(B)$ $\mathrm{N}_2$	$(q)$ Undergoes oxidation
$(C)$ $\mathrm{O}_2^{-}$	$(r)$ Undergoes reduction
$(D)$ $\mathrm{O}_2$	$(s)$ Bond order $\geq 2$
	$(t)$ Mixing of ' $\mathrm{s}$ ' and ' $\mathrm{p}$ ' orbitals

$List-I$	$List-II$
$(A)$ $\Psi_{ MO }=\Psi_{ A }-\Psi_{ B }$	$(I)$ Dipole moment
$(B)$ $\mu=Q \times I$	$(II)$ Bonding molecular orbital
$(C)$ $\frac{N_{b}-N_{a}}{2}$	$(III)$ Anti-bonding molecualr orbital
$(D)$ $\Psi_{ MO }=\Psi_{ A }+\Psi_{ B }$	$(IV)$ Bond order

The bond order of $NO$ molecule is

Solution

Similar Questions

Bond order is a concept in the molecular orbital theory. It depends on the number of electrons in the bonding and antibonding orbitals. Which of the following statements is true about it ? The bond order

Match List $I$ with List $II$ : List $I$ (Molecule Species) List $I$ ( Property Shape ) $A$ $\mathrm{SO}_2 \mathrm{Cl}_2$ $I$ Paramagnetic $B$ $NO$ $II$ Diamagnetic $C$ $\mathrm{NO}_2^{-}$ $III$ Tetrahedral $D$ $\mathrm{I}_3^{-}$ $IV$ Linear Choose the correct answer from the options given below :

In which pair, or pairs, is the stronger bond found in the first species $(a)$ $O_2^ – ,{O_2}$ $(b)$ ${N_2},N_2^ +$ $(c)$ $NO^+, NO^-$

Match $List-I$ with $List-II$. $List-I$ $List-II$ $(A)$ $\Psi_{ MO }=\Psi_{ A }-\Psi_{ B }$ $(I)$ Dipole moment $(B)$ $\mu=Q \times I$ $(II)$ Bonding molecular orbital $(C)$ $\frac{N_{b}-N_{a}}{2}$ $(III)$ Anti-bonding molecualr orbital $(D)$ $\Psi_{ MO }=\Psi_{ A }+\Psi_{ B }$ $(IV)$ Bond order

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In which pair, or pairs, is the stronger bond found in the first species

$(a)$ $O_2^ – ,{O_2}$ $(b)$ ${N_2},N_2^ +$ $(c)$ $NO^+, NO^-$