QBManager

Classification of Elements and Periodicity in Properties

89574 The second ionization enthalpies of $Li, Be, B$ and $C$ are in order

1

Li > B > C > Be

2 Li

> C > B > Be

3 Be

> C > B > Li

4 B

> C > Be > Li

Explanation:

The ionization enthalpy depend on the following factors:
(i) Half- filled or full filled atomic orbitals are more stable than other incomplete orbital.
(ii) Smaller the size of atom/ion, more is the ionization energy. (I.E.) For $2^{nd}$ I.E. all elements have $M^{+}$ configuration

| | $c T o t a l n o . o f | | - - - | - - - | | e l e c t r o n s$ | $c O u t e r m o s t | | c o n f i g u r a t i o n$ |
| $^{+} (= 3)$ | 2 | $1^{2}$ (fully filled) |
| $^{+} (= 4)$ | 3 | $l 2 s^{1} (h a s m o r e s i z e) | | t h a n c a r b o n)$ |
| $^{+} (= 5)$ | 4 | $2^{2}$ (fully filled) |
| $C^{+} (Z = 6)$ | 5 | $2 s^{2} 2 p^{1}$ (Smaller size) |
|
Hence, correct, order for $2^{nd}$ I.E. is
${Li}^{+} > B^{+} > C^{+} > {Be}^{+}$
or $Li > B > C > {Be}^{+}$

CG PET -2017

Classification of Elements and Periodicity in Properties

89582 Identify the elements $X$ and $Y$ using the ionisation energy values given below.
{lll}
| | $l I o n i s a t i o n | | - - - | - - - | | e n e r g y (I s t)$ | $l / | | (I I n d)$ |
| $X$ | 495 | 4563 |
|$$ | 731 | 1450 |
|

1

X = Na, Y = Mg

2

X = Mg, Y = F

3

X = Mg, Y = Na

4

X = F, Y = Mg

Explanation:

The electronic configuration of $Na = [Ne] 3 s^{1}$ ionisatin energy of sodium is very low but second ionisation energy is very high due to stable noble gas configuration of ${Na}^{+}$ .
The electronic configuration of $Mg = [Ne] 3 s^{2}$ first and second ionisation energy is low.

(JEE Main 2021

Classification of Elements and Periodicity in Properties

89585 Choose the correct option regarding the following statements :
Statement-1: Nitrogen has lesser electron gain enthalpy than oxygen.
Statement-2: Oxygen has lesser ionization enthalpy than nitrogen.

1 Statement - 1 is correct. Statement -2 is wrong

2 Both statement 1 and 2 are wrong

3 Both statement 1 and 2 are correct

4 Statement -1 is wrong. Statement -2 is correct.

Explanation:

Electronic configuration of $_{7} N = 1 s^{2}, 2 s^{2}, 2 p^{3}$

Electronic configuration of $_{8} O = 1 s^{2}, 2 s^{2}, 2 p^{4}$

$\begin{array}{lll} ↑↓ & ↑ & ↑ \end{array}$
$N$ has lesser electron gain enthalpy because the extra addition of one electron in $2 p$ orbital is very energy consuming.
Oxygen has lower ionisation enthalpy than nitrogen because by removing one electron from $2 p$ - orbital, oxygen acquires stable configuration i.e. $2 p^{3}$ .On the other hand, in case nitrogen it is not easy to remove one of the three $2 p$ electronic due to its stable configuration.

Shift-II

Classification of Elements and Periodicity in Properties

89586 Which order among the following is incorrect?

1

{NH}_{3} < {PH}_{3} < {AsH}_{3}

: (Acidic nature)

2

Li < Be < B < C :

IE

_{1} (Δ H_{1})

3

{Al}_{2} O_{3} < MgO < {Na}_{2} O < K_{2} O

: (Basic Nature)

4

{Li}^{+} < K^{+} < {Cs}^{+}

: (Ionic radius)

Explanation:

Ionization enthalpy is defined as the amount of energy an isolated gaseous atom. Atom would take to lose an electron in its ground state. So the correct order is-
$\begin{aligned} Li < B < Be < C \\ 520 801 899 1086 kJ / Mole (Δ H_{1}) \end{aligned}$

Shift-I

Classification of Elements and Periodicity in Properties

89574 The second ionization enthalpies of $Li, Be, B$ and $C$ are in order

1

Li > B > C > Be

2 Li

> C > B > Be

3 Be

> C > B > Li

4 B

> C > Be > Li

Explanation:

The ionization enthalpy depend on the following factors:
(i) Half- filled or full filled atomic orbitals are more stable than other incomplete orbital.
(ii) Smaller the size of atom/ion, more is the ionization energy. (I.E.) For $2^{nd}$ I.E. all elements have $M^{+}$ configuration

| | $c T o t a l n o . o f | | - - - | - - - | | e l e c t r o n s$ | $c O u t e r m o s t | | c o n f i g u r a t i o n$ |
| $^{+} (= 3)$ | 2 | $1^{2}$ (fully filled) |
| $^{+} (= 4)$ | 3 | $l 2 s^{1} (h a s m o r e s i z e) | | t h a n c a r b o n)$ |
| $^{+} (= 5)$ | 4 | $2^{2}$ (fully filled) |
| $C^{+} (Z = 6)$ | 5 | $2 s^{2} 2 p^{1}$ (Smaller size) |
|
Hence, correct, order for $2^{nd}$ I.E. is
${Li}^{+} > B^{+} > C^{+} > {Be}^{+}$
or $Li > B > C > {Be}^{+}$

CG PET -2017

Classification of Elements and Periodicity in Properties

89582 Identify the elements $X$ and $Y$ using the ionisation energy values given below.
{lll}
| | $l I o n i s a t i o n | | - - - | - - - | | e n e r g y (I s t)$ | $l / | | (I I n d)$ |
| $X$ | 495 | 4563 |
|$$ | 731 | 1450 |
|

1

X = Na, Y = Mg

2

X = Mg, Y = F

3

X = Mg, Y = Na

4

X = F, Y = Mg

Explanation:

The electronic configuration of $Na = [Ne] 3 s^{1}$ ionisatin energy of sodium is very low but second ionisation energy is very high due to stable noble gas configuration of ${Na}^{+}$ .
The electronic configuration of $Mg = [Ne] 3 s^{2}$ first and second ionisation energy is low.

(JEE Main 2021

Classification of Elements and Periodicity in Properties

89585 Choose the correct option regarding the following statements :
Statement-1: Nitrogen has lesser electron gain enthalpy than oxygen.
Statement-2: Oxygen has lesser ionization enthalpy than nitrogen.

1 Statement - 1 is correct. Statement -2 is wrong

2 Both statement 1 and 2 are wrong

3 Both statement 1 and 2 are correct

4 Statement -1 is wrong. Statement -2 is correct.

Explanation:

Electronic configuration of $_{7} N = 1 s^{2}, 2 s^{2}, 2 p^{3}$

Electronic configuration of $_{8} O = 1 s^{2}, 2 s^{2}, 2 p^{4}$

$\begin{array}{lll} ↑↓ & ↑ & ↑ \end{array}$
$N$ has lesser electron gain enthalpy because the extra addition of one electron in $2 p$ orbital is very energy consuming.
Oxygen has lower ionisation enthalpy than nitrogen because by removing one electron from $2 p$ - orbital, oxygen acquires stable configuration i.e. $2 p^{3}$ .On the other hand, in case nitrogen it is not easy to remove one of the three $2 p$ electronic due to its stable configuration.

Shift-II

Classification of Elements and Periodicity in Properties

89586 Which order among the following is incorrect?

1

{NH}_{3} < {PH}_{3} < {AsH}_{3}

: (Acidic nature)

2

Li < Be < B < C :

IE

_{1} (Δ H_{1})

3

{Al}_{2} O_{3} < MgO < {Na}_{2} O < K_{2} O

: (Basic Nature)

4

{Li}^{+} < K^{+} < {Cs}^{+}

: (Ionic radius)

Explanation:

Ionization enthalpy is defined as the amount of energy an isolated gaseous atom. Atom would take to lose an electron in its ground state. So the correct order is-
$\begin{aligned} Li < B < Be < C \\ 520 801 899 1086 kJ / Mole (Δ H_{1}) \end{aligned}$

Shift-I

Classification of Elements and Periodicity in Properties

89574 The second ionization enthalpies of $Li, Be, B$ and $C$ are in order

1

Li > B > C > Be

2 Li

> C > B > Be

3 Be

> C > B > Li

4 B

> C > Be > Li

Explanation:

The ionization enthalpy depend on the following factors:
(i) Half- filled or full filled atomic orbitals are more stable than other incomplete orbital.
(ii) Smaller the size of atom/ion, more is the ionization energy. (I.E.) For $2^{nd}$ I.E. all elements have $M^{+}$ configuration

| | $c T o t a l n o . o f | | - - - | - - - | | e l e c t r o n s$ | $c O u t e r m o s t | | c o n f i g u r a t i o n$ |
| $^{+} (= 3)$ | 2 | $1^{2}$ (fully filled) |
| $^{+} (= 4)$ | 3 | $l 2 s^{1} (h a s m o r e s i z e) | | t h a n c a r b o n)$ |
| $^{+} (= 5)$ | 4 | $2^{2}$ (fully filled) |
| $C^{+} (Z = 6)$ | 5 | $2 s^{2} 2 p^{1}$ (Smaller size) |
|
Hence, correct, order for $2^{nd}$ I.E. is
${Li}^{+} > B^{+} > C^{+} > {Be}^{+}$
or $Li > B > C > {Be}^{+}$

CG PET -2017

Classification of Elements and Periodicity in Properties

89582 Identify the elements $X$ and $Y$ using the ionisation energy values given below.
{lll}
| | $l I o n i s a t i o n | | - - - | - - - | | e n e r g y (I s t)$ | $l / | | (I I n d)$ |
| $X$ | 495 | 4563 |
|$$ | 731 | 1450 |
|

1

X = Na, Y = Mg

2

X = Mg, Y = F

3

X = Mg, Y = Na

4

X = F, Y = Mg

Explanation:

The electronic configuration of $Na = [Ne] 3 s^{1}$ ionisatin energy of sodium is very low but second ionisation energy is very high due to stable noble gas configuration of ${Na}^{+}$ .
The electronic configuration of $Mg = [Ne] 3 s^{2}$ first and second ionisation energy is low.

(JEE Main 2021

Classification of Elements and Periodicity in Properties

89585 Choose the correct option regarding the following statements :
Statement-1: Nitrogen has lesser electron gain enthalpy than oxygen.
Statement-2: Oxygen has lesser ionization enthalpy than nitrogen.

1 Statement - 1 is correct. Statement -2 is wrong

2 Both statement 1 and 2 are wrong

3 Both statement 1 and 2 are correct

4 Statement -1 is wrong. Statement -2 is correct.

Explanation:

Electronic configuration of $_{7} N = 1 s^{2}, 2 s^{2}, 2 p^{3}$

Electronic configuration of $_{8} O = 1 s^{2}, 2 s^{2}, 2 p^{4}$

$\begin{array}{lll} ↑↓ & ↑ & ↑ \end{array}$
$N$ has lesser electron gain enthalpy because the extra addition of one electron in $2 p$ orbital is very energy consuming.
Oxygen has lower ionisation enthalpy than nitrogen because by removing one electron from $2 p$ - orbital, oxygen acquires stable configuration i.e. $2 p^{3}$ .On the other hand, in case nitrogen it is not easy to remove one of the three $2 p$ electronic due to its stable configuration.

Shift-II

Classification of Elements and Periodicity in Properties

89586 Which order among the following is incorrect?

1

{NH}_{3} < {PH}_{3} < {AsH}_{3}

: (Acidic nature)

2

Li < Be < B < C :

IE

_{1} (Δ H_{1})

3

{Al}_{2} O_{3} < MgO < {Na}_{2} O < K_{2} O

: (Basic Nature)

4

{Li}^{+} < K^{+} < {Cs}^{+}

: (Ionic radius)

Explanation:

Ionization enthalpy is defined as the amount of energy an isolated gaseous atom. Atom would take to lose an electron in its ground state. So the correct order is-
$\begin{aligned} Li < B < Be < C \\ 520 801 899 1086 kJ / Mole (Δ H_{1}) \end{aligned}$

Shift-I

Classification of Elements and Periodicity in Properties

89574 The second ionization enthalpies of $Li, Be, B$ and $C$ are in order

1

Li > B > C > Be

2 Li

> C > B > Be

3 Be

> C > B > Li

4 B

> C > Be > Li

Explanation:

The ionization enthalpy depend on the following factors:
(i) Half- filled or full filled atomic orbitals are more stable than other incomplete orbital.
(ii) Smaller the size of atom/ion, more is the ionization energy. (I.E.) For $2^{nd}$ I.E. all elements have $M^{+}$ configuration

| | $c T o t a l n o . o f | | - - - | - - - | | e l e c t r o n s$ | $c O u t e r m o s t | | c o n f i g u r a t i o n$ |
| $^{+} (= 3)$ | 2 | $1^{2}$ (fully filled) |
| $^{+} (= 4)$ | 3 | $l 2 s^{1} (h a s m o r e s i z e) | | t h a n c a r b o n)$ |
| $^{+} (= 5)$ | 4 | $2^{2}$ (fully filled) |
| $C^{+} (Z = 6)$ | 5 | $2 s^{2} 2 p^{1}$ (Smaller size) |
|
Hence, correct, order for $2^{nd}$ I.E. is
${Li}^{+} > B^{+} > C^{+} > {Be}^{+}$
or $Li > B > C > {Be}^{+}$

CG PET -2017

Classification of Elements and Periodicity in Properties

89582 Identify the elements $X$ and $Y$ using the ionisation energy values given below.
{lll}
| | $l I o n i s a t i o n | | - - - | - - - | | e n e r g y (I s t)$ | $l / | | (I I n d)$ |
| $X$ | 495 | 4563 |
|$$ | 731 | 1450 |
|

1

X = Na, Y = Mg

2

X = Mg, Y = F

3

X = Mg, Y = Na

4

X = F, Y = Mg

Explanation:

The electronic configuration of $Na = [Ne] 3 s^{1}$ ionisatin energy of sodium is very low but second ionisation energy is very high due to stable noble gas configuration of ${Na}^{+}$ .
The electronic configuration of $Mg = [Ne] 3 s^{2}$ first and second ionisation energy is low.

(JEE Main 2021

Classification of Elements and Periodicity in Properties

89585 Choose the correct option regarding the following statements :
Statement-1: Nitrogen has lesser electron gain enthalpy than oxygen.
Statement-2: Oxygen has lesser ionization enthalpy than nitrogen.

1 Statement - 1 is correct. Statement -2 is wrong

2 Both statement 1 and 2 are wrong

3 Both statement 1 and 2 are correct

4 Statement -1 is wrong. Statement -2 is correct.

Explanation:

Electronic configuration of $_{7} N = 1 s^{2}, 2 s^{2}, 2 p^{3}$

Electronic configuration of $_{8} O = 1 s^{2}, 2 s^{2}, 2 p^{4}$

$\begin{array}{lll} ↑↓ & ↑ & ↑ \end{array}$
$N$ has lesser electron gain enthalpy because the extra addition of one electron in $2 p$ orbital is very energy consuming.
Oxygen has lower ionisation enthalpy than nitrogen because by removing one electron from $2 p$ - orbital, oxygen acquires stable configuration i.e. $2 p^{3}$ .On the other hand, in case nitrogen it is not easy to remove one of the three $2 p$ electronic due to its stable configuration.

Shift-II

Classification of Elements and Periodicity in Properties

89586 Which order among the following is incorrect?

1

{NH}_{3} < {PH}_{3} < {AsH}_{3}

: (Acidic nature)

2

Li < Be < B < C :

IE

_{1} (Δ H_{1})

3

{Al}_{2} O_{3} < MgO < {Na}_{2} O < K_{2} O

: (Basic Nature)

4

{Li}^{+} < K^{+} < {Cs}^{+}

: (Ionic radius)

Explanation:

Ionization enthalpy is defined as the amount of energy an isolated gaseous atom. Atom would take to lose an electron in its ground state. So the correct order is-
$\begin{aligned} Li < B < Be < C \\ 520 801 899 1086 kJ / Mole (Δ H_{1}) \end{aligned}$

Shift-I

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