An element X is having the first ionisation enthalpy as 496kJ/mol. It is having second ionisation enthalpy as 4564kJ/mol. While it is having third ionisation enthalpy as 6918kJ/mol. And accordingly the fourth ionisation enthalpy is 9542kJ/mol. Find the formula of the sulphate and chloride of the element.

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Solution

1^st ionization energy = 496kJ/mol.
2^nd ionization energy= 4564kJ/mol.
3^rd ionization energy = 6918kJ/mol.
4^th ionization energy = 9542kJ/mol.

Here the difference of 1^st and 2^nd ionization energy is very high it means the given element tends to lose one electron and after that it must get some stable configuration so that it needs very high amount of energy for next electron.
Hence element M can form M⁺¹ ion easily and formulae will be as given below;

Chloride = MCl

Sulphate = M₂SO₄

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Similar questions

X

Y

	1st Ionisation Enthalpy(KJ/mol)	2nd Ionisation Enthalpy(KJ/mol)
$X$	$495$	$4563$
$Y$	$731$	$1450$

N H_{4} O H

C H_{3} C O O H

m o l^{- 1}

C H_{3} C O O H

m o l^{- 1}

N H_{4} O H

O_{2} (g)

O^{2 -} (g)

639 {kJ mol}^{- 1}

O^{-} (g)

141 {kJ mol}^{- 1}

(Δ_{i} H_{1} k J m o l^{- 1})

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