$A B$ is predominately ionic as $A^{+}$ $B^{-}$ if:

(I P)_{A} < (I P)_{B}

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(E A)_{A} < (E A)_{B}

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(E N)_{A} < (E N)_{B}

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(I P)_{B} < (I P)_{A}

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Solution

The correct option is B $(E N)_{A} < (E N)_{B}$
$A \to A^{+} + e^{-}$ if $(I P)_{A} < (I P)_{B}$

$B + e^{-} \to B^{-}$ if $(E A)_{B} > (E A)_{A}$

$(E N) = \frac{(I P) + (E A)}{2}$

$A^{+} B^{-}$ is formed

if $(E N)_{A} < (E N)_{B}$

Hence, the correct option is $C$

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AB is predominately ionic as A+ B− if:

The correct option is B (EN)A<(EN)BA→A++e− if (IP)A<(IP)BB+e−→B− if (EA)B>(EA)A(EN)=(IP)+(EA)2A+B− is formedif (EN)A<(EN)BHence, the correct option is C

$A B$ is predominately ionic as $A^{+}$ $B^{-}$ if:

The correct option is B $(E N)_{A} < (E N)_{B}$
$A \to A^{+} + e^{-}$ if $(I P)_{A} < (I P)_{B}$

$B + e^{-} \to B^{-}$ if $(E A)_{B} > (E A)_{A}$

$(E N) = \frac{(I P) + (E A)}{2}$

$A^{+} B^{-}$ is formed

if $(E N)_{A} < (E N)_{B}$

Hence, the correct option is $C$