QBManager

Classification of Elements and Periodicity in Properties

89412 The correct sequence which shows decreasing order of the ionic radii of the elements is

1

{Al}^{3 +} > {Mg}^{2 +} > {Na}^{+} > F^{-} > O^{2 -}

2

{Na}^{+} > {Mg}^{2 +} > {Al}^{3 +} > O^{2 -} > F^{-}

3

{Na}^{+} > F^{-} > {Mg}^{2 +} > O^{2 -} > {Al}^{3 +}

4

O^{2 -} > F^{-} > {Na}^{+} > {Mg}^{2 +} > {Al}^{3 +}

Explanation:

$\frac{Z}{e} = \frac{nuclear charge}{no. of electrons}$
When $Z / e$ ratio decrease the size increase for
${\begin{cases} O^{2 -}, \frac{Z}{e} = \frac{8}{10} = 0.8 \\ F^{-}, \frac{Z}{e} = \frac{9}{10} = 0.9 \\ {Na}^{+}, \frac{Z}{e} = \frac{11}{10} = 1.1 \\ {Mg}^{2 +}, \frac{Z}{e} = \frac{12}{10} = 1.2 \\ {Al}^{3 +}, \frac{Z}{e} = \frac{13}{10} = 1.3 \end{cases}$
Hence, the decreasing order of ionic radii of the element are-
$O^{2 -} > F^{-} > {Na}^{+} > {Mg}^{2 +} > {Al}^{3 +}$

JEE Main-04.09.2020

Classification of Elements and Periodicity in Properties

89418 The set representing the correct order of ionic radius is

1

{Li}^{+} > {Be}^{2 +} > {Na}^{+} > {Mg}^{2 +}

2

{Na}^{+} > {Li}^{+} > {Mg}^{2 +} > {Be}^{2 +}

3

{Li}^{2 +} > {Na}^{+} > {Mg}^{2 +} > {Be}^{2 +}

4

{Mg}^{2 +} > {Be}^{2 +} > {Li}^{+} > {Na}^{+}

Explanation:

Ionic radii increases when moving down the group and its decrease as the positive charge on the element increases due to effective nuclear charge. In the cation, number of electrons are less than number of protons, hence its effective nuclear charge increase and ionic radii decrease.
The correct order of ionic radii-
${Na}_{95}^{+} > \underset{76}{{Li}^{+}} > \underset{72}{{Mg}^{2 +}} > {Be}_{31}^{2 +}$
Ionic radii $(pm)$

UPTU/UPSEE-2015

Classification of Elements and Periodicity in Properties

89452 What is the correct increasing order of ionic or atomic radii in the following?

1

{Si}^{4 +} < P^{5 +} < S^{6 +} < {Cl}^{7 +}

2

P^{5 +} < {Si}^{4 +} < {Cl}^{7 +} < S^{6 +}

3

{Cl}^{7 +} < S^{6 +} < P^{5 +} < {Si}^{4 +}

4

S^{6 +} < P^{5 +} < {Cl}^{7 +} < {Si}^{4 +}

Explanation:

For
$\begin{aligned} Si = 14, {Si}^{4 +} = 10 e^{-} \\ P = 15, P^{+ 5} = 10 e^{-} \\ S = 16, S^{+ 6} = 10 e^{-} \\ Cl = 17, {Cl}^{+ 7} = 10 e^{-} \end{aligned}$
Ionic radii $\propto \frac{1}{charge}$
Greater the charge ionic radii decrease
Hence, the increasing order of ionic radii are-
${Cl}^{7 +} < S^{6 +} < P^{5 +} < {Si}^{4 +}$

J and K CET-(2018)

Classification of Elements and Periodicity in Properties

89445 Find the correct order of the given elements with respect to their size

1

Zn > Fe > {Fe}^{2 +} > {Fe}^{3 +}

2

{Fe}^{2 +} > {Fe}^{3 +} > Zn > Fe

3

Fe > {Fe}^{2 +} > {Fe}^{3 +} > Zn

4

Zn > {Fe}^{3 +} > {Fe}^{2 +} > Fe

Explanation:

$Zn$ atomic radii is $137 pm$ .
$Fe$ atomic radii is $126 pm$ .
${Fe}^{3 +}$ ionic radius is $63 pm$ .
${Fe}^{2 +}$ has an ionic radius of $77 pm$ .
Thus, the correct order of their size is-
$Zn > Fe > {Fe}^{2 +} > {Fe}^{3 +}$
Hence, the correct option is (a).

Shift-II

Classification of Elements and Periodicity in Properties

89449 Which has least covalent radius?

1

Mn

2

Cu

3

Zn

4

Ni

Explanation:

Covalent radius of an atom denotes half of the internuclear distance between the atoms of the monoatomic species.
It is different from ionic and atomic radius. It depends on the bond order of the compound formed by the covalent bonding.
Hence, the covalent radius in $Pm$ are-
$Ni < Cu < Zn = Mn$
$\begin{array}{llll} 125 & 128 & 137 & 137 \end{array}$
So, the least covalent radius is $Ni$ .

AIIMS 26 May 2019 (Morning)

Classification of Elements and Periodicity in Properties

89412 The correct sequence which shows decreasing order of the ionic radii of the elements is

1

{Al}^{3 +} > {Mg}^{2 +} > {Na}^{+} > F^{-} > O^{2 -}

2

{Na}^{+} > {Mg}^{2 +} > {Al}^{3 +} > O^{2 -} > F^{-}

3

{Na}^{+} > F^{-} > {Mg}^{2 +} > O^{2 -} > {Al}^{3 +}

4

O^{2 -} > F^{-} > {Na}^{+} > {Mg}^{2 +} > {Al}^{3 +}

Explanation:

$\frac{Z}{e} = \frac{nuclear charge}{no. of electrons}$
When $Z / e$ ratio decrease the size increase for
${\begin{cases} O^{2 -}, \frac{Z}{e} = \frac{8}{10} = 0.8 \\ F^{-}, \frac{Z}{e} = \frac{9}{10} = 0.9 \\ {Na}^{+}, \frac{Z}{e} = \frac{11}{10} = 1.1 \\ {Mg}^{2 +}, \frac{Z}{e} = \frac{12}{10} = 1.2 \\ {Al}^{3 +}, \frac{Z}{e} = \frac{13}{10} = 1.3 \end{cases}$
Hence, the decreasing order of ionic radii of the element are-
$O^{2 -} > F^{-} > {Na}^{+} > {Mg}^{2 +} > {Al}^{3 +}$

JEE Main-04.09.2020

Classification of Elements and Periodicity in Properties

89418 The set representing the correct order of ionic radius is

1

{Li}^{+} > {Be}^{2 +} > {Na}^{+} > {Mg}^{2 +}

2

{Na}^{+} > {Li}^{+} > {Mg}^{2 +} > {Be}^{2 +}

3

{Li}^{2 +} > {Na}^{+} > {Mg}^{2 +} > {Be}^{2 +}

4

{Mg}^{2 +} > {Be}^{2 +} > {Li}^{+} > {Na}^{+}

Explanation:

Ionic radii increases when moving down the group and its decrease as the positive charge on the element increases due to effective nuclear charge. In the cation, number of electrons are less than number of protons, hence its effective nuclear charge increase and ionic radii decrease.
The correct order of ionic radii-
${Na}_{95}^{+} > \underset{76}{{Li}^{+}} > \underset{72}{{Mg}^{2 +}} > {Be}_{31}^{2 +}$
Ionic radii $(pm)$

UPTU/UPSEE-2015

Classification of Elements and Periodicity in Properties

89452 What is the correct increasing order of ionic or atomic radii in the following?

1

{Si}^{4 +} < P^{5 +} < S^{6 +} < {Cl}^{7 +}

2

P^{5 +} < {Si}^{4 +} < {Cl}^{7 +} < S^{6 +}

3

{Cl}^{7 +} < S^{6 +} < P^{5 +} < {Si}^{4 +}

4

S^{6 +} < P^{5 +} < {Cl}^{7 +} < {Si}^{4 +}

Explanation:

For
$\begin{aligned} Si = 14, {Si}^{4 +} = 10 e^{-} \\ P = 15, P^{+ 5} = 10 e^{-} \\ S = 16, S^{+ 6} = 10 e^{-} \\ Cl = 17, {Cl}^{+ 7} = 10 e^{-} \end{aligned}$
Ionic radii $\propto \frac{1}{charge}$
Greater the charge ionic radii decrease
Hence, the increasing order of ionic radii are-
${Cl}^{7 +} < S^{6 +} < P^{5 +} < {Si}^{4 +}$

J and K CET-(2018)

Classification of Elements and Periodicity in Properties

89445 Find the correct order of the given elements with respect to their size

1

Zn > Fe > {Fe}^{2 +} > {Fe}^{3 +}

2

{Fe}^{2 +} > {Fe}^{3 +} > Zn > Fe

3

Fe > {Fe}^{2 +} > {Fe}^{3 +} > Zn

4

Zn > {Fe}^{3 +} > {Fe}^{2 +} > Fe

Explanation:

$Zn$ atomic radii is $137 pm$ .
$Fe$ atomic radii is $126 pm$ .
${Fe}^{3 +}$ ionic radius is $63 pm$ .
${Fe}^{2 +}$ has an ionic radius of $77 pm$ .
Thus, the correct order of their size is-
$Zn > Fe > {Fe}^{2 +} > {Fe}^{3 +}$
Hence, the correct option is (a).

Shift-II

Classification of Elements and Periodicity in Properties

89449 Which has least covalent radius?

1

Mn

2

Cu

3

Zn

4

Ni

Explanation:

Covalent radius of an atom denotes half of the internuclear distance between the atoms of the monoatomic species.
It is different from ionic and atomic radius. It depends on the bond order of the compound formed by the covalent bonding.
Hence, the covalent radius in $Pm$ are-
$Ni < Cu < Zn = Mn$
$\begin{array}{llll} 125 & 128 & 137 & 137 \end{array}$
So, the least covalent radius is $Ni$ .

AIIMS 26 May 2019 (Morning)

Classification of Elements and Periodicity in Properties

89412 The correct sequence which shows decreasing order of the ionic radii of the elements is

1

{Al}^{3 +} > {Mg}^{2 +} > {Na}^{+} > F^{-} > O^{2 -}

2

{Na}^{+} > {Mg}^{2 +} > {Al}^{3 +} > O^{2 -} > F^{-}

3

{Na}^{+} > F^{-} > {Mg}^{2 +} > O^{2 -} > {Al}^{3 +}

4

O^{2 -} > F^{-} > {Na}^{+} > {Mg}^{2 +} > {Al}^{3 +}

Explanation:

$\frac{Z}{e} = \frac{nuclear charge}{no. of electrons}$
When $Z / e$ ratio decrease the size increase for
${\begin{cases} O^{2 -}, \frac{Z}{e} = \frac{8}{10} = 0.8 \\ F^{-}, \frac{Z}{e} = \frac{9}{10} = 0.9 \\ {Na}^{+}, \frac{Z}{e} = \frac{11}{10} = 1.1 \\ {Mg}^{2 +}, \frac{Z}{e} = \frac{12}{10} = 1.2 \\ {Al}^{3 +}, \frac{Z}{e} = \frac{13}{10} = 1.3 \end{cases}$
Hence, the decreasing order of ionic radii of the element are-
$O^{2 -} > F^{-} > {Na}^{+} > {Mg}^{2 +} > {Al}^{3 +}$

JEE Main-04.09.2020

Classification of Elements and Periodicity in Properties

89418 The set representing the correct order of ionic radius is

1

{Li}^{+} > {Be}^{2 +} > {Na}^{+} > {Mg}^{2 +}

2

{Na}^{+} > {Li}^{+} > {Mg}^{2 +} > {Be}^{2 +}

3

{Li}^{2 +} > {Na}^{+} > {Mg}^{2 +} > {Be}^{2 +}

4

{Mg}^{2 +} > {Be}^{2 +} > {Li}^{+} > {Na}^{+}

Explanation:

Ionic radii increases when moving down the group and its decrease as the positive charge on the element increases due to effective nuclear charge. In the cation, number of electrons are less than number of protons, hence its effective nuclear charge increase and ionic radii decrease.
The correct order of ionic radii-
${Na}_{95}^{+} > \underset{76}{{Li}^{+}} > \underset{72}{{Mg}^{2 +}} > {Be}_{31}^{2 +}$
Ionic radii $(pm)$

UPTU/UPSEE-2015

Classification of Elements and Periodicity in Properties

89452 What is the correct increasing order of ionic or atomic radii in the following?

1

{Si}^{4 +} < P^{5 +} < S^{6 +} < {Cl}^{7 +}

2

P^{5 +} < {Si}^{4 +} < {Cl}^{7 +} < S^{6 +}

3

{Cl}^{7 +} < S^{6 +} < P^{5 +} < {Si}^{4 +}

4

S^{6 +} < P^{5 +} < {Cl}^{7 +} < {Si}^{4 +}

Explanation:

For
$\begin{aligned} Si = 14, {Si}^{4 +} = 10 e^{-} \\ P = 15, P^{+ 5} = 10 e^{-} \\ S = 16, S^{+ 6} = 10 e^{-} \\ Cl = 17, {Cl}^{+ 7} = 10 e^{-} \end{aligned}$
Ionic radii $\propto \frac{1}{charge}$
Greater the charge ionic radii decrease
Hence, the increasing order of ionic radii are-
${Cl}^{7 +} < S^{6 +} < P^{5 +} < {Si}^{4 +}$

J and K CET-(2018)

Classification of Elements and Periodicity in Properties

89445 Find the correct order of the given elements with respect to their size

1

Zn > Fe > {Fe}^{2 +} > {Fe}^{3 +}

2

{Fe}^{2 +} > {Fe}^{3 +} > Zn > Fe

3

Fe > {Fe}^{2 +} > {Fe}^{3 +} > Zn

4

Zn > {Fe}^{3 +} > {Fe}^{2 +} > Fe

Explanation:

$Zn$ atomic radii is $137 pm$ .
$Fe$ atomic radii is $126 pm$ .
${Fe}^{3 +}$ ionic radius is $63 pm$ .
${Fe}^{2 +}$ has an ionic radius of $77 pm$ .
Thus, the correct order of their size is-
$Zn > Fe > {Fe}^{2 +} > {Fe}^{3 +}$
Hence, the correct option is (a).

Shift-II

Classification of Elements and Periodicity in Properties

89449 Which has least covalent radius?

1

Mn

2

Cu

3

Zn

4

Ni

Explanation:

Covalent radius of an atom denotes half of the internuclear distance between the atoms of the monoatomic species.
It is different from ionic and atomic radius. It depends on the bond order of the compound formed by the covalent bonding.
Hence, the covalent radius in $Pm$ are-
$Ni < Cu < Zn = Mn$
$\begin{array}{llll} 125 & 128 & 137 & 137 \end{array}$
So, the least covalent radius is $Ni$ .

AIIMS 26 May 2019 (Morning)

Classification of Elements and Periodicity in Properties

89412 The correct sequence which shows decreasing order of the ionic radii of the elements is

1

{Al}^{3 +} > {Mg}^{2 +} > {Na}^{+} > F^{-} > O^{2 -}

2

{Na}^{+} > {Mg}^{2 +} > {Al}^{3 +} > O^{2 -} > F^{-}

3

{Na}^{+} > F^{-} > {Mg}^{2 +} > O^{2 -} > {Al}^{3 +}

4

O^{2 -} > F^{-} > {Na}^{+} > {Mg}^{2 +} > {Al}^{3 +}

Explanation:

$\frac{Z}{e} = \frac{nuclear charge}{no. of electrons}$
When $Z / e$ ratio decrease the size increase for
${\begin{cases} O^{2 -}, \frac{Z}{e} = \frac{8}{10} = 0.8 \\ F^{-}, \frac{Z}{e} = \frac{9}{10} = 0.9 \\ {Na}^{+}, \frac{Z}{e} = \frac{11}{10} = 1.1 \\ {Mg}^{2 +}, \frac{Z}{e} = \frac{12}{10} = 1.2 \\ {Al}^{3 +}, \frac{Z}{e} = \frac{13}{10} = 1.3 \end{cases}$
Hence, the decreasing order of ionic radii of the element are-
$O^{2 -} > F^{-} > {Na}^{+} > {Mg}^{2 +} > {Al}^{3 +}$

JEE Main-04.09.2020

Classification of Elements and Periodicity in Properties

89418 The set representing the correct order of ionic radius is

1

{Li}^{+} > {Be}^{2 +} > {Na}^{+} > {Mg}^{2 +}

2

{Na}^{+} > {Li}^{+} > {Mg}^{2 +} > {Be}^{2 +}

3

{Li}^{2 +} > {Na}^{+} > {Mg}^{2 +} > {Be}^{2 +}

4

{Mg}^{2 +} > {Be}^{2 +} > {Li}^{+} > {Na}^{+}

Explanation:

Ionic radii increases when moving down the group and its decrease as the positive charge on the element increases due to effective nuclear charge. In the cation, number of electrons are less than number of protons, hence its effective nuclear charge increase and ionic radii decrease.
The correct order of ionic radii-
${Na}_{95}^{+} > \underset{76}{{Li}^{+}} > \underset{72}{{Mg}^{2 +}} > {Be}_{31}^{2 +}$
Ionic radii $(pm)$

UPTU/UPSEE-2015

Classification of Elements and Periodicity in Properties

89452 What is the correct increasing order of ionic or atomic radii in the following?

1

{Si}^{4 +} < P^{5 +} < S^{6 +} < {Cl}^{7 +}

2

P^{5 +} < {Si}^{4 +} < {Cl}^{7 +} < S^{6 +}

3

{Cl}^{7 +} < S^{6 +} < P^{5 +} < {Si}^{4 +}

4

S^{6 +} < P^{5 +} < {Cl}^{7 +} < {Si}^{4 +}

Explanation:

For
$\begin{aligned} Si = 14, {Si}^{4 +} = 10 e^{-} \\ P = 15, P^{+ 5} = 10 e^{-} \\ S = 16, S^{+ 6} = 10 e^{-} \\ Cl = 17, {Cl}^{+ 7} = 10 e^{-} \end{aligned}$
Ionic radii $\propto \frac{1}{charge}$
Greater the charge ionic radii decrease
Hence, the increasing order of ionic radii are-
${Cl}^{7 +} < S^{6 +} < P^{5 +} < {Si}^{4 +}$

J and K CET-(2018)

Classification of Elements and Periodicity in Properties

89445 Find the correct order of the given elements with respect to their size

1

Zn > Fe > {Fe}^{2 +} > {Fe}^{3 +}

2

{Fe}^{2 +} > {Fe}^{3 +} > Zn > Fe

3

Fe > {Fe}^{2 +} > {Fe}^{3 +} > Zn

4

Zn > {Fe}^{3 +} > {Fe}^{2 +} > Fe

Explanation:

$Zn$ atomic radii is $137 pm$ .
$Fe$ atomic radii is $126 pm$ .
${Fe}^{3 +}$ ionic radius is $63 pm$ .
${Fe}^{2 +}$ has an ionic radius of $77 pm$ .
Thus, the correct order of their size is-
$Zn > Fe > {Fe}^{2 +} > {Fe}^{3 +}$
Hence, the correct option is (a).

Shift-II

Classification of Elements and Periodicity in Properties

89449 Which has least covalent radius?

1

Mn

2

Cu

3

Zn

4

Ni

Explanation:

Covalent radius of an atom denotes half of the internuclear distance between the atoms of the monoatomic species.
It is different from ionic and atomic radius. It depends on the bond order of the compound formed by the covalent bonding.
Hence, the covalent radius in $Pm$ are-
$Ni < Cu < Zn = Mn$
$\begin{array}{llll} 125 & 128 & 137 & 137 \end{array}$
So, the least covalent radius is $Ni$ .

AIIMS 26 May 2019 (Morning)

Classification of Elements and Periodicity in Properties

89412 The correct sequence which shows decreasing order of the ionic radii of the elements is

1

{Al}^{3 +} > {Mg}^{2 +} > {Na}^{+} > F^{-} > O^{2 -}

2

{Na}^{+} > {Mg}^{2 +} > {Al}^{3 +} > O^{2 -} > F^{-}

3

{Na}^{+} > F^{-} > {Mg}^{2 +} > O^{2 -} > {Al}^{3 +}

4

O^{2 -} > F^{-} > {Na}^{+} > {Mg}^{2 +} > {Al}^{3 +}

Explanation:

$\frac{Z}{e} = \frac{nuclear charge}{no. of electrons}$
When $Z / e$ ratio decrease the size increase for
${\begin{cases} O^{2 -}, \frac{Z}{e} = \frac{8}{10} = 0.8 \\ F^{-}, \frac{Z}{e} = \frac{9}{10} = 0.9 \\ {Na}^{+}, \frac{Z}{e} = \frac{11}{10} = 1.1 \\ {Mg}^{2 +}, \frac{Z}{e} = \frac{12}{10} = 1.2 \\ {Al}^{3 +}, \frac{Z}{e} = \frac{13}{10} = 1.3 \end{cases}$
Hence, the decreasing order of ionic radii of the element are-
$O^{2 -} > F^{-} > {Na}^{+} > {Mg}^{2 +} > {Al}^{3 +}$

JEE Main-04.09.2020

Classification of Elements and Periodicity in Properties

89418 The set representing the correct order of ionic radius is

1

{Li}^{+} > {Be}^{2 +} > {Na}^{+} > {Mg}^{2 +}

2

{Na}^{+} > {Li}^{+} > {Mg}^{2 +} > {Be}^{2 +}

3

{Li}^{2 +} > {Na}^{+} > {Mg}^{2 +} > {Be}^{2 +}

4

{Mg}^{2 +} > {Be}^{2 +} > {Li}^{+} > {Na}^{+}

Explanation:

Ionic radii increases when moving down the group and its decrease as the positive charge on the element increases due to effective nuclear charge. In the cation, number of electrons are less than number of protons, hence its effective nuclear charge increase and ionic radii decrease.
The correct order of ionic radii-
${Na}_{95}^{+} > \underset{76}{{Li}^{+}} > \underset{72}{{Mg}^{2 +}} > {Be}_{31}^{2 +}$
Ionic radii $(pm)$

UPTU/UPSEE-2015

Classification of Elements and Periodicity in Properties

89452 What is the correct increasing order of ionic or atomic radii in the following?

1

{Si}^{4 +} < P^{5 +} < S^{6 +} < {Cl}^{7 +}

2

P^{5 +} < {Si}^{4 +} < {Cl}^{7 +} < S^{6 +}

3

{Cl}^{7 +} < S^{6 +} < P^{5 +} < {Si}^{4 +}

4

S^{6 +} < P^{5 +} < {Cl}^{7 +} < {Si}^{4 +}

Explanation:

For
$\begin{aligned} Si = 14, {Si}^{4 +} = 10 e^{-} \\ P = 15, P^{+ 5} = 10 e^{-} \\ S = 16, S^{+ 6} = 10 e^{-} \\ Cl = 17, {Cl}^{+ 7} = 10 e^{-} \end{aligned}$
Ionic radii $\propto \frac{1}{charge}$
Greater the charge ionic radii decrease
Hence, the increasing order of ionic radii are-
${Cl}^{7 +} < S^{6 +} < P^{5 +} < {Si}^{4 +}$

J and K CET-(2018)

Classification of Elements and Periodicity in Properties

89445 Find the correct order of the given elements with respect to their size

1

Zn > Fe > {Fe}^{2 +} > {Fe}^{3 +}

2

{Fe}^{2 +} > {Fe}^{3 +} > Zn > Fe

3

Fe > {Fe}^{2 +} > {Fe}^{3 +} > Zn

4

Zn > {Fe}^{3 +} > {Fe}^{2 +} > Fe

Explanation:

$Zn$ atomic radii is $137 pm$ .
$Fe$ atomic radii is $126 pm$ .
${Fe}^{3 +}$ ionic radius is $63 pm$ .
${Fe}^{2 +}$ has an ionic radius of $77 pm$ .
Thus, the correct order of their size is-
$Zn > Fe > {Fe}^{2 +} > {Fe}^{3 +}$
Hence, the correct option is (a).

Shift-II

Classification of Elements and Periodicity in Properties

89449 Which has least covalent radius?

1

Mn

2

Cu

3

Zn

4

Ni

Explanation:

Covalent radius of an atom denotes half of the internuclear distance between the atoms of the monoatomic species.
It is different from ionic and atomic radius. It depends on the bond order of the compound formed by the covalent bonding.
Hence, the covalent radius in $Pm$ are-
$Ni < Cu < Zn = Mn$
$\begin{array}{llll} 125 & 128 & 137 & 137 \end{array}$
So, the least covalent radius is $Ni$ .

AIIMS 26 May 2019 (Morning)