QBManager

CHXI03:CLASSIFICATION OF ELEMENTS AND PERIODICITY IN PROPERTIES

313247 The correct increasing order of the ionic radii is

1

C a^{2 +} < K^{+} < C l^{-} < S^{2 -}

2

K^{+} < S^{2 -} < C a^{2 +} < C l^{-}

3

S^{2 -} < C l^{-} < C a^{2 +} < K^{+}

4

C l^{-} < C a^{2 +} < K^{+} < S^{2 -}

Explanation:

In case of isoelectronic species, as the – ve charge increases radii increases and as the + ve charge increases radii decreases. Hence, the correct order is $C a^{2 +} < K^{+} < C l^{-} < S^{2 -} .$

JEE - 2023

CHXI03:CLASSIFICATION OF ELEMENTS AND PERIODICITY IN PROPERTIES

313248 Assertion :
The ionic size of ${Mg}^{2 +} > {Al}^{3 +}$ .
Reason :
In isoelectronic species, greater the nuclear charge, less is the size.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Both Assertion and Reason are incorrect

Explanation:

The ionic size of ${Mg}^{2 +}$ is larger than ${Al}^{3 +}$ . Both the species are isoelectronic and in case of isoelectronic species as nuclear charge increases, attractive force between electron of valence shell and nucleus increases which results in decrease in ionic size.
Hence, both Assertion and Reason are correct and Reason is the correct explanation of Assertion.

AIIMS - 2008

CHXI03:CLASSIFICATION OF ELEMENTS AND PERIODICITY IN PROPERTIES

313249 Increasing order of ionic size for the ions, $F^{-}, O^{2 -}, {Na}^{+}, {Al}^{3 +}$ is

1

O^{2 -} < F^{-} < {Na}^{+} < {Al}^{3 +}

2

{Al}^{3 +} < {Na}^{+} < F^{-} < O^{2 -}

3

O^{2 -} < {Na}^{+} < F^{-} < {Al}^{3 +}

4

{Al}^{3 +} < F^{-} < {Na}^{+} < O^{2 -}

Explanation:

For isoelectronic species,
atomic size $\propto \frac{1}{atomic number}$
Thus, the correct order would be
$\underset{(13)}{{Al}^{3 +}} < \underset{(11)}{{Na}^{+}} < \underset{(9)}{F^{-}} < \underset{(8)}{O^{2 -}}$

CHXI03:CLASSIFICATION OF ELEMENTS AND PERIODICITY IN PROPERTIES

313250 The correct arrangement for the ions in the increasing order of their radii is

1

N a^{+}, C l^{-}, C a^{2 +}

2

C a^{2 +}, K^{+}, S^{2 -}

3

N a^{+}, A l^{3 +}, B e^{2 +}

4

C l^{-}, F^{-}, S^{2 -}

Explanation:

supporting img
The species $C a^{2 +}, K^{+} a n d S^{2 -}$ are isoelec tronic ions, the radius decreases as the nuclear charge increases. Hence, the increasing order of their radii is: $C a^{2 +} < K^{+} < S^{2 -}$

KCET - 2014

CHXI03:CLASSIFICATION OF ELEMENTS AND PERIODICITY IN PROPERTIES

313251 Which one of the following ions has the highest value of ionic radius?

1

O^{2 -}

2

B^{3 +}

3

{Li}^{+}

4

F^{-}

Explanation:

$O^{2 -}$ has the highest value of ionic radius and this can be explained on the basis of Z/e (nuclear charge/ number of electrons) ratio. When $Z / e$ ratio increases, the size decreases and when vice-versa $\frac{Z}{e} \propto \frac{1}{r}$

CHXI03:CLASSIFICATION OF ELEMENTS AND PERIODICITY IN PROPERTIES

313247 The correct increasing order of the ionic radii is

1

C a^{2 +} < K^{+} < C l^{-} < S^{2 -}

2

K^{+} < S^{2 -} < C a^{2 +} < C l^{-}

3

S^{2 -} < C l^{-} < C a^{2 +} < K^{+}

4

C l^{-} < C a^{2 +} < K^{+} < S^{2 -}

Explanation:

In case of isoelectronic species, as the – ve charge increases radii increases and as the + ve charge increases radii decreases. Hence, the correct order is $C a^{2 +} < K^{+} < C l^{-} < S^{2 -} .$

JEE - 2023

CHXI03:CLASSIFICATION OF ELEMENTS AND PERIODICITY IN PROPERTIES

313248 Assertion :
The ionic size of ${Mg}^{2 +} > {Al}^{3 +}$ .
Reason :
In isoelectronic species, greater the nuclear charge, less is the size.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Both Assertion and Reason are incorrect

Explanation:

The ionic size of ${Mg}^{2 +}$ is larger than ${Al}^{3 +}$ . Both the species are isoelectronic and in case of isoelectronic species as nuclear charge increases, attractive force between electron of valence shell and nucleus increases which results in decrease in ionic size.
Hence, both Assertion and Reason are correct and Reason is the correct explanation of Assertion.

AIIMS - 2008

CHXI03:CLASSIFICATION OF ELEMENTS AND PERIODICITY IN PROPERTIES

313249 Increasing order of ionic size for the ions, $F^{-}, O^{2 -}, {Na}^{+}, {Al}^{3 +}$ is

1

O^{2 -} < F^{-} < {Na}^{+} < {Al}^{3 +}

2

{Al}^{3 +} < {Na}^{+} < F^{-} < O^{2 -}

3

O^{2 -} < {Na}^{+} < F^{-} < {Al}^{3 +}

4

{Al}^{3 +} < F^{-} < {Na}^{+} < O^{2 -}

Explanation:

For isoelectronic species,
atomic size $\propto \frac{1}{atomic number}$
Thus, the correct order would be
$\underset{(13)}{{Al}^{3 +}} < \underset{(11)}{{Na}^{+}} < \underset{(9)}{F^{-}} < \underset{(8)}{O^{2 -}}$

CHXI03:CLASSIFICATION OF ELEMENTS AND PERIODICITY IN PROPERTIES

313250 The correct arrangement for the ions in the increasing order of their radii is

1

N a^{+}, C l^{-}, C a^{2 +}

2

C a^{2 +}, K^{+}, S^{2 -}

3

N a^{+}, A l^{3 +}, B e^{2 +}

4

C l^{-}, F^{-}, S^{2 -}

Explanation:

supporting img
The species $C a^{2 +}, K^{+} a n d S^{2 -}$ are isoelec tronic ions, the radius decreases as the nuclear charge increases. Hence, the increasing order of their radii is: $C a^{2 +} < K^{+} < S^{2 -}$

KCET - 2014

CHXI03:CLASSIFICATION OF ELEMENTS AND PERIODICITY IN PROPERTIES

313251 Which one of the following ions has the highest value of ionic radius?

1

O^{2 -}

2

B^{3 +}

3

{Li}^{+}

4

F^{-}

Explanation:

$O^{2 -}$ has the highest value of ionic radius and this can be explained on the basis of Z/e (nuclear charge/ number of electrons) ratio. When $Z / e$ ratio increases, the size decreases and when vice-versa $\frac{Z}{e} \propto \frac{1}{r}$

CHXI03:CLASSIFICATION OF ELEMENTS AND PERIODICITY IN PROPERTIES

313247 The correct increasing order of the ionic radii is

1

C a^{2 +} < K^{+} < C l^{-} < S^{2 -}

2

K^{+} < S^{2 -} < C a^{2 +} < C l^{-}

3

S^{2 -} < C l^{-} < C a^{2 +} < K^{+}

4

C l^{-} < C a^{2 +} < K^{+} < S^{2 -}

Explanation:

In case of isoelectronic species, as the – ve charge increases radii increases and as the + ve charge increases radii decreases. Hence, the correct order is $C a^{2 +} < K^{+} < C l^{-} < S^{2 -} .$

JEE - 2023

CHXI03:CLASSIFICATION OF ELEMENTS AND PERIODICITY IN PROPERTIES

313248 Assertion :
The ionic size of ${Mg}^{2 +} > {Al}^{3 +}$ .
Reason :
In isoelectronic species, greater the nuclear charge, less is the size.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Both Assertion and Reason are incorrect

Explanation:

The ionic size of ${Mg}^{2 +}$ is larger than ${Al}^{3 +}$ . Both the species are isoelectronic and in case of isoelectronic species as nuclear charge increases, attractive force between electron of valence shell and nucleus increases which results in decrease in ionic size.
Hence, both Assertion and Reason are correct and Reason is the correct explanation of Assertion.

AIIMS - 2008

CHXI03:CLASSIFICATION OF ELEMENTS AND PERIODICITY IN PROPERTIES

313249 Increasing order of ionic size for the ions, $F^{-}, O^{2 -}, {Na}^{+}, {Al}^{3 +}$ is

1

O^{2 -} < F^{-} < {Na}^{+} < {Al}^{3 +}

2

{Al}^{3 +} < {Na}^{+} < F^{-} < O^{2 -}

3

O^{2 -} < {Na}^{+} < F^{-} < {Al}^{3 +}

4

{Al}^{3 +} < F^{-} < {Na}^{+} < O^{2 -}

Explanation:

For isoelectronic species,
atomic size $\propto \frac{1}{atomic number}$
Thus, the correct order would be
$\underset{(13)}{{Al}^{3 +}} < \underset{(11)}{{Na}^{+}} < \underset{(9)}{F^{-}} < \underset{(8)}{O^{2 -}}$

CHXI03:CLASSIFICATION OF ELEMENTS AND PERIODICITY IN PROPERTIES

313250 The correct arrangement for the ions in the increasing order of their radii is

1

N a^{+}, C l^{-}, C a^{2 +}

2

C a^{2 +}, K^{+}, S^{2 -}

3

N a^{+}, A l^{3 +}, B e^{2 +}

4

C l^{-}, F^{-}, S^{2 -}

Explanation:

supporting img
The species $C a^{2 +}, K^{+} a n d S^{2 -}$ are isoelec tronic ions, the radius decreases as the nuclear charge increases. Hence, the increasing order of their radii is: $C a^{2 +} < K^{+} < S^{2 -}$

KCET - 2014

CHXI03:CLASSIFICATION OF ELEMENTS AND PERIODICITY IN PROPERTIES

313251 Which one of the following ions has the highest value of ionic radius?

1

O^{2 -}

2

B^{3 +}

3

{Li}^{+}

4

F^{-}

Explanation:

$O^{2 -}$ has the highest value of ionic radius and this can be explained on the basis of Z/e (nuclear charge/ number of electrons) ratio. When $Z / e$ ratio increases, the size decreases and when vice-versa $\frac{Z}{e} \propto \frac{1}{r}$

CHXI03:CLASSIFICATION OF ELEMENTS AND PERIODICITY IN PROPERTIES

313247 The correct increasing order of the ionic radii is

1

C a^{2 +} < K^{+} < C l^{-} < S^{2 -}

2

K^{+} < S^{2 -} < C a^{2 +} < C l^{-}

3

S^{2 -} < C l^{-} < C a^{2 +} < K^{+}

4

C l^{-} < C a^{2 +} < K^{+} < S^{2 -}

Explanation:

In case of isoelectronic species, as the – ve charge increases radii increases and as the + ve charge increases radii decreases. Hence, the correct order is $C a^{2 +} < K^{+} < C l^{-} < S^{2 -} .$

JEE - 2023

CHXI03:CLASSIFICATION OF ELEMENTS AND PERIODICITY IN PROPERTIES

313248 Assertion :
The ionic size of ${Mg}^{2 +} > {Al}^{3 +}$ .
Reason :
In isoelectronic species, greater the nuclear charge, less is the size.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Both Assertion and Reason are incorrect

Explanation:

The ionic size of ${Mg}^{2 +}$ is larger than ${Al}^{3 +}$ . Both the species are isoelectronic and in case of isoelectronic species as nuclear charge increases, attractive force between electron of valence shell and nucleus increases which results in decrease in ionic size.
Hence, both Assertion and Reason are correct and Reason is the correct explanation of Assertion.

AIIMS - 2008

CHXI03:CLASSIFICATION OF ELEMENTS AND PERIODICITY IN PROPERTIES

313249 Increasing order of ionic size for the ions, $F^{-}, O^{2 -}, {Na}^{+}, {Al}^{3 +}$ is

1

O^{2 -} < F^{-} < {Na}^{+} < {Al}^{3 +}

2

{Al}^{3 +} < {Na}^{+} < F^{-} < O^{2 -}

3

O^{2 -} < {Na}^{+} < F^{-} < {Al}^{3 +}

4

{Al}^{3 +} < F^{-} < {Na}^{+} < O^{2 -}

Explanation:

For isoelectronic species,
atomic size $\propto \frac{1}{atomic number}$
Thus, the correct order would be
$\underset{(13)}{{Al}^{3 +}} < \underset{(11)}{{Na}^{+}} < \underset{(9)}{F^{-}} < \underset{(8)}{O^{2 -}}$

CHXI03:CLASSIFICATION OF ELEMENTS AND PERIODICITY IN PROPERTIES

313250 The correct arrangement for the ions in the increasing order of their radii is

1

N a^{+}, C l^{-}, C a^{2 +}

2

C a^{2 +}, K^{+}, S^{2 -}

3

N a^{+}, A l^{3 +}, B e^{2 +}

4

C l^{-}, F^{-}, S^{2 -}

Explanation:

supporting img
The species $C a^{2 +}, K^{+} a n d S^{2 -}$ are isoelec tronic ions, the radius decreases as the nuclear charge increases. Hence, the increasing order of their radii is: $C a^{2 +} < K^{+} < S^{2 -}$

KCET - 2014

CHXI03:CLASSIFICATION OF ELEMENTS AND PERIODICITY IN PROPERTIES

313251 Which one of the following ions has the highest value of ionic radius?

1

O^{2 -}

2

B^{3 +}

3

{Li}^{+}

4

F^{-}

Explanation:

$O^{2 -}$ has the highest value of ionic radius and this can be explained on the basis of Z/e (nuclear charge/ number of electrons) ratio. When $Z / e$ ratio increases, the size decreases and when vice-versa $\frac{Z}{e} \propto \frac{1}{r}$

CHXI03:CLASSIFICATION OF ELEMENTS AND PERIODICITY IN PROPERTIES

313247 The correct increasing order of the ionic radii is

1

C a^{2 +} < K^{+} < C l^{-} < S^{2 -}

2

K^{+} < S^{2 -} < C a^{2 +} < C l^{-}

3

S^{2 -} < C l^{-} < C a^{2 +} < K^{+}

4

C l^{-} < C a^{2 +} < K^{+} < S^{2 -}

Explanation:

In case of isoelectronic species, as the – ve charge increases radii increases and as the + ve charge increases radii decreases. Hence, the correct order is $C a^{2 +} < K^{+} < C l^{-} < S^{2 -} .$

JEE - 2023

CHXI03:CLASSIFICATION OF ELEMENTS AND PERIODICITY IN PROPERTIES

313248 Assertion :
The ionic size of ${Mg}^{2 +} > {Al}^{3 +}$ .
Reason :
In isoelectronic species, greater the nuclear charge, less is the size.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Both Assertion and Reason are incorrect

Explanation:

The ionic size of ${Mg}^{2 +}$ is larger than ${Al}^{3 +}$ . Both the species are isoelectronic and in case of isoelectronic species as nuclear charge increases, attractive force between electron of valence shell and nucleus increases which results in decrease in ionic size.
Hence, both Assertion and Reason are correct and Reason is the correct explanation of Assertion.

AIIMS - 2008

CHXI03:CLASSIFICATION OF ELEMENTS AND PERIODICITY IN PROPERTIES

313249 Increasing order of ionic size for the ions, $F^{-}, O^{2 -}, {Na}^{+}, {Al}^{3 +}$ is

1

O^{2 -} < F^{-} < {Na}^{+} < {Al}^{3 +}

2

{Al}^{3 +} < {Na}^{+} < F^{-} < O^{2 -}

3

O^{2 -} < {Na}^{+} < F^{-} < {Al}^{3 +}

4

{Al}^{3 +} < F^{-} < {Na}^{+} < O^{2 -}

Explanation:

For isoelectronic species,
atomic size $\propto \frac{1}{atomic number}$
Thus, the correct order would be
$\underset{(13)}{{Al}^{3 +}} < \underset{(11)}{{Na}^{+}} < \underset{(9)}{F^{-}} < \underset{(8)}{O^{2 -}}$

CHXI03:CLASSIFICATION OF ELEMENTS AND PERIODICITY IN PROPERTIES

313250 The correct arrangement for the ions in the increasing order of their radii is

1

N a^{+}, C l^{-}, C a^{2 +}

2

C a^{2 +}, K^{+}, S^{2 -}

3

N a^{+}, A l^{3 +}, B e^{2 +}

4

C l^{-}, F^{-}, S^{2 -}

Explanation:

supporting img
The species $C a^{2 +}, K^{+} a n d S^{2 -}$ are isoelec tronic ions, the radius decreases as the nuclear charge increases. Hence, the increasing order of their radii is: $C a^{2 +} < K^{+} < S^{2 -}$

KCET - 2014

CHXI03:CLASSIFICATION OF ELEMENTS AND PERIODICITY IN PROPERTIES

313251 Which one of the following ions has the highest value of ionic radius?

1

O^{2 -}

2

B^{3 +}

3

{Li}^{+}

4

F^{-}

Explanation:

$O^{2 -}$ has the highest value of ionic radius and this can be explained on the basis of Z/e (nuclear charge/ number of electrons) ratio. When $Z / e$ ratio increases, the size decreases and when vice-versa $\frac{Z}{e} \propto \frac{1}{r}$