The first ionization enthalpy of Na , Mg and Si, respectively, are: 496, 737 and 786 kJ mol -1. The first ionization enthalpy ( kJ mol -1) of Al is:

Question

The first ionization enthalpy of Na , Mg and Si, respectively, are: 496, 737 and 786 kJ mol -1. The first ionization enthalpy ( kJ mol -1) of Al is: (A) 487 (B) 768 (C) 577 (D) 856

Tardigrade Admin · Accepted Answer

I. E: Na < Al < Mg < Si ∴ 496< IE ( Al )<737 Option (C), matches the condition. i.e textIE( Al )=577 kJ mol -1

Q. The first ionization enthalpy of $N a, M g$ and Si, respectively, are: $496, 737$ and $786 k J m o l^{- 1}$ . The first ionization enthalpy $(k J m o l^{- 1})$ of $A l$ is:

487

768

577

856

I. E : $N a < A l < M g < S i$
$∴ 496 < I E (A l) < 737$
Option (C), matches the condition.
i.e $IE (A l) = 577 k J m o l^{- 1}$

Q. The first ionization enthalpy of Na,Mg and Si, respectively, are: 496,737 and 786kJmol−1. The first ionization enthalpy (kJmol−1) of Al is:

487

768

577

856

I. E : Na<Al<Mg<Si ∴496<IE(Al)<737 Option (C), matches the condition. i.e IE(Al)=577kJmol−1

Q. The first ionization enthalpy of $N a, M g$ and Si, respectively, are: $496, 737$ and $786 k J m o l^{- 1}$ . The first ionization enthalpy $(k J m o l^{- 1})$ of $A l$ is:

I. E : $N a < A l < M g < S i$
$∴ 496 < I E (A l) < 737$
Option (C), matches the condition.
i.e $IE (A l) = 577 k J m o l^{- 1}$