Question

Increasing order of effective nuclear charge in na, al, mg and si atoms

Answer 1

Answer:

Na<Mg>Al<Si

Explanation:

• The Ionisation energy of My will be larger than of Na due to fully filled configuration (3s)2
• the ionisation of Al will be smaller than that of Mg due to one electron extra than the stable configuration . But smaller than Si due to increase in effective nuclear charge of Si
• =Na<Mg>Al<Si

Answer 2

Note: The correct question must be provided with following options-

The increasing order of effective nuclear charge in $Na, Al, Mg$ and $Si$ atoms-

a) $Na < Mg < Si < Al$

b) $Na < Mg < Al < Si$

c) $Mg < Na < Al < Si$

d) $Na = Mg = Al = Si$

Explanation for correct option

b:

1. The atomic number increases from left to right across a period, as does the effective nuclear charge.
2. The valence electrons in a group retain the same effective nuclear charge, but the orbitals are now further from the nucleus.
3. As a result, the nucleus exerts less pull on the outer electrons, and the atomic radii are larger.
4. Because sodium and magnesium are in the same period, the outer electrons of the $Mg$ atom have a higher effective nuclear charge, and the magnesium cation,  $Mg^{2+}$, has a higher nuclear charge than the  $Na^{+}$, cation.
5. An aluminium ion typically has a charge of $3+$.
6. Silicon has a higher nuclear charge $4+$ than aluminium.
7. As a result, in the case of silicon, the unpaired '3p' electron will experience more effective nuclear charge.

Thus, the most appropriate order of effective nuclear energy is$Na < Mg < Al < Si$.

Explanation for incorrect options

a:

Silicon is more nuclearly charged than aluminium.

Thus, option a is incorrect.

c:

Magnesium has higher nuclear charge than Sodium.

Thus, option c is incorrect.

d:

None of the effective nuclear charge on $Na, Al, Mg, Si$ is equal.

Thus, option d is also incorrect.

Hence, option b) is correct.

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