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JAMB Chemistry 2019 Paper

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Question : 54

Total: 80

The electronic configuration of element Z is 1s

^{2}

^{2}

^{6}

^{2}

^{1}

. What is the formula of the compound formed between Z and tetraoxosulphate (VI) ion.

Z $_{3}$ (SO $_{4}$ ) $_{2}$
ZSO $_{4}$
Z $_{2}$ SO $_{4}$
Z $_{2}$ (SO $_{4}$ ) $_{3}$

Solution:

Concept:

The element forms a cation by losing its valence electrons, and the formula is obtained by balancing charges with the sulphate ion.

Explanation:

The configuration

1s^{2}2s^{2}2p^{6}3s^{2}3p^{1}

shows that Z has 3 electrons in its outermost shell (

3s^{2}3p^{1}

).

Therefore, Z loses 3 electrons to form the cation

Z^{3+}

.

Tetraoxosulphate (VI) ion is

SO_{4}^{2-}

.

To balance charges, the criss-cross method gives

Z_{2}(SO_{4})_{3}

.

Thus, the correct formula is

Z_{2}(SO_{4})_{3}

.

Answer:

D.

Z_{2}(SO_{4})_{3}

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