In a binary compound formed by two elements A and B is given by the formula AB. If (IP)A+(EA)A5.6>(IP)B+(EA)B5.6, where [IP and EA are in eV per atom] then A– B bond.

EN in Mullikan scale =(1P)A+(EA)A5.6more electronegative atom gets negative plurality

In a binary compound formed by two elements A and B is given by the formula AB. If (IP)A+(EA)A5.6>(IP)B+(EA)B5.6, where [IP and EA are in eV per atom] then A– B bond.

EN in Mullikan scale =(1P)A+(EA)A5.6more electronegative atom gets negative plurality

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In a binary compound formed by two elements A and B is given by the formula AB. If $\frac{{(I P)}_{A} + {(E A)}_{A}}{5.6} > \frac{{(I P)}_{B} + {(E A)}_{B}}{5.6},$ where [IP and EA are in eV per atom] then A– B bond.

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can be polar covalent of the type $\overset{δ +}{A} - \overset{δ -}{B}$

is purely ionic bond

can be polar covalent of the type $\overset{δ -}{A} - \overset{δ +}{B}$

is purely covalent

answer is C.

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$E_{N}$ in Mullikan scale $= \frac{{(1 P)}_{A} + {(E A)}_{A}}{5.6}$

more electronegative atom gets negative plurality

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In a binary compound formed by two elements A and B is given by the formula AB. If (IP)A+(EA)A5.6>(IP)B+(EA)B5.6, where [IP and EA are in eV per atom] then A– B bond.

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In a binary compound formed by two elements A and B is given by the formula AB. If $\frac{{(I P)}_{A} + {(E A)}_{A}}{5.6} > \frac{{(I P)}_{B} + {(E A)}_{B}}{5.6},$ where [IP and EA are in eV per atom] then A– B bond.