Question

Elements X,Y, Z have atomic numbers 6, 9 and 12 respectively. Which one:

(i) forms anions - negative ions;

(ii) form cations - positive ions;

iii) has 4 electrons in the outermost orbit ?

Answer 1

Electronic configuration: Electronic configuration is the arrangement of electrons in the increasing order of energy of orbits in an atom.

Electronic configuration of X, Y, Z:

${\mathrm{X}}_6=\underset{\mathrm{K}}{2},\underset{\mathrm{L}}{4}$

$Y_9=\underset{\mathrm{K}}{2},\underset{\mathrm{L}}{7}$

$Z_{12}=\underset{\mathrm{K}}{2},\underset{\mathrm{L}}{8},\underset{\mathrm{M}}{2}$

Part (i): Form anions - negative ion

• An atom that has an incomplete valence shell and can attain noble gas configuration form an anion by gaining electrons.
• Y will form an anion (negative ion) because it will gain one electron to achieve inert gas configuration and will form ${\mathrm{Y}}^{-}$ anion.

Part (ii): Form cations - positive ion

• An atom that has an incomplete valence shell and can attain noble gas configuration form a cation by losing electrons.
• Z will form a cation (positive ion) because it will donate two electrons to achieve an inert gas configuration and will form $Z^{++}$ cation.

Part (iii): four electrons in the outermost orbit

• X is having 4 electrons in its outermost orbit which will share 4 electrons to achieve inert gas configuration.
• Electronic configuration of ${\mathrm{X}}_6=\underset{\mathrm{K}}{2},\underset{\mathrm{L}}{4}$