Classification of Elements and Periodicity in Properties
89514
Which has the smallest size? 4. Electro Negativity
1 $\mathrm{Na}^{+}$
2 $\mathrm{Mg}^{2+}$
3 $\mathrm{Al}^{3+}$
4 $\mathrm{P}^{5+}$
Explanation:
The smallest size is $\mathrm{P}^{5+}$ because having maximum nuclear charge per electron. When nuclear charge increases then atomic size decreases.
JEE Main
Classification of Elements and Periodicity in Properties
89429
Assertion (A) : Beryllium and aluminum forms $\operatorname{Be} F_4^{2-}$ and $\mathrm{Al} F_6^{3-}$ Reason(R) : Both $\mathbf{B}^{2+}$ and $\mathrm{Al}^{3+}$ have almost same ionic radius The correct option among the following is
1 (A) is true, (R) is true and (R) is the correct explanation for (A)
2 (A) is true, (R) is true but (R) is not the correct explanation for (A)
3 (A) is true but (R) is false
4 (A) is false but (R) is true
Explanation:
Beryllium and aluminum forms $\mathrm{BeF}_4^{-2}$ and $\mathrm{AlF}_6^{3-}$ and $\mathrm{Be}^{+2}$ and $\mathrm{Al}^{+3}$ have almost same ionic radius due to about same electronegativity, polarizing power and charge/radius radio, of their ion.
Classification of Elements and Periodicity in Properties
89514
Which has the smallest size? 4. Electro Negativity
1 $\mathrm{Na}^{+}$
2 $\mathrm{Mg}^{2+}$
3 $\mathrm{Al}^{3+}$
4 $\mathrm{P}^{5+}$
Explanation:
The smallest size is $\mathrm{P}^{5+}$ because having maximum nuclear charge per electron. When nuclear charge increases then atomic size decreases.
JEE Main
Classification of Elements and Periodicity in Properties
89429
Assertion (A) : Beryllium and aluminum forms $\operatorname{Be} F_4^{2-}$ and $\mathrm{Al} F_6^{3-}$ Reason(R) : Both $\mathbf{B}^{2+}$ and $\mathrm{Al}^{3+}$ have almost same ionic radius The correct option among the following is
1 (A) is true, (R) is true and (R) is the correct explanation for (A)
2 (A) is true, (R) is true but (R) is not the correct explanation for (A)
3 (A) is true but (R) is false
4 (A) is false but (R) is true
Explanation:
Beryllium and aluminum forms $\mathrm{BeF}_4^{-2}$ and $\mathrm{AlF}_6^{3-}$ and $\mathrm{Be}^{+2}$ and $\mathrm{Al}^{+3}$ have almost same ionic radius due to about same electronegativity, polarizing power and charge/radius radio, of their ion.