QBManager

Classification of Elements and Periodicity in Properties

89521 Which of the following is not true for oxidation?

1 Addition of oxygen

2 addition of electronegative element

3 removal of hydrogen

4 removal of electronegative element

Kerala-CEE-29.08.2021

Classification of Elements and Periodicity in Properties

89518 Which of the following indicates the correct variation in electronegativities?

1 F

>

N

<

O

> C

2 F

>

O

> N > C

3 F

<

N

<

O

< C

4

F > N > O < C

Explanation:

On moving from left to right in a $2^{nd}$ period electronegativity increases.
$\begin{array}{lllll} Element & C & N & O & F \\ EN, Value & 2.5 & 3.0 & 3.5 & 4 \end{array}$
Hence, the order of $EN$ are-
$F > O > N > C$

CG PET - 2018

Classification of Elements and Periodicity in Properties

89519 Pauling's equation for determining the electronegativity of an element is

1

X_{A} - X_{B} = 0.208 \sqrt{Δ}

2

X_{A} + X_{B} = 0.208 \sqrt{Δ}

3

X_{A} - X_{B} = 0.208 Δ^{2}

4

X_{A} - X_{B} = \sqrt{Δ}

Explanation:

According to Pauiling equation.
$\begin{aligned} X_{A} - X_{B} = 0.208 {[E_{A - B} - {(E_{A - A} \times E_{B - B})}^{1 / 2}]}^{1 / 2} \\ X_{A} - X_{B} = 0.208 \sqrt{Δ} \end{aligned}$
Where, $Δ$ is the difference of bond dissociation energies of $A - A$ and B-B bonds.
$X_{A}$ and $X_{B}$ are the electronegativities of $A$ and $B$ respectively.
The factor 0.208 arises from the conversion of Kcals to electron volt.

J and K CET-(2006)

Classification of Elements and Periodicity in Properties

89526 The electronegativity of the given elements increases in the order

1 C, N, Si, P

2 N, Si, C, P

3

Si, P, C, N

4 P, Si, N, C

Explanation:

Electro negativity of given element is-
$\begin{aligned} Si = 1.9 \\ P = 2.19 \\ C = 2.55 \\ N = 3.04 \end{aligned}$
Therefore the increasing order is $- Si < P < C < N$ .

AP EAPCET 20.08.2021 Shift-II

Classification of Elements and Periodicity in Properties

89528 Which pair of elements has maximum electro negativity difference?

1 Li \& F

2

Na & F

3

Na & Br

4

Na & Cl

Explanation:

Among the following $Li$ and $Na$ are the most electropositive and $F, Cl$ and $Br$ are most electronegative.
So, the electronegative of the species are-
| Atom | $F$ | $C l$ | $B r$ | $I$ | $L i$ | $N a$ |
|---|---|---|---|---|---|---|
|EN Value | 4 | 3 | 2.8 | 2.5 | 1 | 0.9 |
|
The difference in electronegativity are.
(i) $F - Li = 4 - 2.5 = 1.5$
(ii) $F - Na = 4 - 0.9 = 3.1$
(iii) $Br - Na = 2.8 - 0.9 = 1.9$
(iv) $Cl - Na = 3 - 0.9 = 2.1$
Hence, the $Na$ and $F$ pairs of elements has maximum electronegativity difference.

AIIMS 26 May 2019 (Evening)

Classification of Elements and Periodicity in Properties

89521 Which of the following is not true for oxidation?

1 Addition of oxygen

2 addition of electronegative element

3 removal of hydrogen

4 removal of electronegative element

Kerala-CEE-29.08.2021

Classification of Elements and Periodicity in Properties

89518 Which of the following indicates the correct variation in electronegativities?

1 F

>

N

<

O

> C

2 F

>

O

> N > C

3 F

<

N

<

O

< C

4

F > N > O < C

Explanation:

On moving from left to right in a $2^{nd}$ period electronegativity increases.
$\begin{array}{lllll} Element & C & N & O & F \\ EN, Value & 2.5 & 3.0 & 3.5 & 4 \end{array}$
Hence, the order of $EN$ are-
$F > O > N > C$

CG PET - 2018

Classification of Elements and Periodicity in Properties

89519 Pauling's equation for determining the electronegativity of an element is

1

X_{A} - X_{B} = 0.208 \sqrt{Δ}

2

X_{A} + X_{B} = 0.208 \sqrt{Δ}

3

X_{A} - X_{B} = 0.208 Δ^{2}

4

X_{A} - X_{B} = \sqrt{Δ}

Explanation:

According to Pauiling equation.
$\begin{aligned} X_{A} - X_{B} = 0.208 {[E_{A - B} - {(E_{A - A} \times E_{B - B})}^{1 / 2}]}^{1 / 2} \\ X_{A} - X_{B} = 0.208 \sqrt{Δ} \end{aligned}$
Where, $Δ$ is the difference of bond dissociation energies of $A - A$ and B-B bonds.
$X_{A}$ and $X_{B}$ are the electronegativities of $A$ and $B$ respectively.
The factor 0.208 arises from the conversion of Kcals to electron volt.

J and K CET-(2006)

Classification of Elements and Periodicity in Properties

89526 The electronegativity of the given elements increases in the order

1 C, N, Si, P

2 N, Si, C, P

3

Si, P, C, N

4 P, Si, N, C

Explanation:

Electro negativity of given element is-
$\begin{aligned} Si = 1.9 \\ P = 2.19 \\ C = 2.55 \\ N = 3.04 \end{aligned}$
Therefore the increasing order is $- Si < P < C < N$ .

AP EAPCET 20.08.2021 Shift-II

Classification of Elements and Periodicity in Properties

89528 Which pair of elements has maximum electro negativity difference?

1 Li \& F

2

Na & F

3

Na & Br

4

Na & Cl

Explanation:

Among the following $Li$ and $Na$ are the most electropositive and $F, Cl$ and $Br$ are most electronegative.
So, the electronegative of the species are-
| Atom | $F$ | $C l$ | $B r$ | $I$ | $L i$ | $N a$ |
|---|---|---|---|---|---|---|
|EN Value | 4 | 3 | 2.8 | 2.5 | 1 | 0.9 |
|
The difference in electronegativity are.
(i) $F - Li = 4 - 2.5 = 1.5$
(ii) $F - Na = 4 - 0.9 = 3.1$
(iii) $Br - Na = 2.8 - 0.9 = 1.9$
(iv) $Cl - Na = 3 - 0.9 = 2.1$
Hence, the $Na$ and $F$ pairs of elements has maximum electronegativity difference.

AIIMS 26 May 2019 (Evening)

Classification of Elements and Periodicity in Properties

89521 Which of the following is not true for oxidation?

1 Addition of oxygen

2 addition of electronegative element

3 removal of hydrogen

4 removal of electronegative element

Kerala-CEE-29.08.2021

Classification of Elements and Periodicity in Properties

89518 Which of the following indicates the correct variation in electronegativities?

1 F

>

N

<

O

> C

2 F

>

O

> N > C

3 F

<

N

<

O

< C

4

F > N > O < C

Explanation:

On moving from left to right in a $2^{nd}$ period electronegativity increases.
$\begin{array}{lllll} Element & C & N & O & F \\ EN, Value & 2.5 & 3.0 & 3.5 & 4 \end{array}$
Hence, the order of $EN$ are-
$F > O > N > C$

CG PET - 2018

Classification of Elements and Periodicity in Properties

89519 Pauling's equation for determining the electronegativity of an element is

1

X_{A} - X_{B} = 0.208 \sqrt{Δ}

2

X_{A} + X_{B} = 0.208 \sqrt{Δ}

3

X_{A} - X_{B} = 0.208 Δ^{2}

4

X_{A} - X_{B} = \sqrt{Δ}

Explanation:

According to Pauiling equation.
$\begin{aligned} X_{A} - X_{B} = 0.208 {[E_{A - B} - {(E_{A - A} \times E_{B - B})}^{1 / 2}]}^{1 / 2} \\ X_{A} - X_{B} = 0.208 \sqrt{Δ} \end{aligned}$
Where, $Δ$ is the difference of bond dissociation energies of $A - A$ and B-B bonds.
$X_{A}$ and $X_{B}$ are the electronegativities of $A$ and $B$ respectively.
The factor 0.208 arises from the conversion of Kcals to electron volt.

J and K CET-(2006)

Classification of Elements and Periodicity in Properties

89526 The electronegativity of the given elements increases in the order

1 C, N, Si, P

2 N, Si, C, P

3

Si, P, C, N

4 P, Si, N, C

Explanation:

Electro negativity of given element is-
$\begin{aligned} Si = 1.9 \\ P = 2.19 \\ C = 2.55 \\ N = 3.04 \end{aligned}$
Therefore the increasing order is $- Si < P < C < N$ .

AP EAPCET 20.08.2021 Shift-II

Classification of Elements and Periodicity in Properties

89528 Which pair of elements has maximum electro negativity difference?

1 Li \& F

2

Na & F

3

Na & Br

4

Na & Cl

Explanation:

Among the following $Li$ and $Na$ are the most electropositive and $F, Cl$ and $Br$ are most electronegative.
So, the electronegative of the species are-
| Atom | $F$ | $C l$ | $B r$ | $I$ | $L i$ | $N a$ |
|---|---|---|---|---|---|---|
|EN Value | 4 | 3 | 2.8 | 2.5 | 1 | 0.9 |
|
The difference in electronegativity are.
(i) $F - Li = 4 - 2.5 = 1.5$
(ii) $F - Na = 4 - 0.9 = 3.1$
(iii) $Br - Na = 2.8 - 0.9 = 1.9$
(iv) $Cl - Na = 3 - 0.9 = 2.1$
Hence, the $Na$ and $F$ pairs of elements has maximum electronegativity difference.

AIIMS 26 May 2019 (Evening)

Classification of Elements and Periodicity in Properties

89521 Which of the following is not true for oxidation?

1 Addition of oxygen

2 addition of electronegative element

3 removal of hydrogen

4 removal of electronegative element

Kerala-CEE-29.08.2021

Classification of Elements and Periodicity in Properties

89518 Which of the following indicates the correct variation in electronegativities?

1 F

>

N

<

O

> C

2 F

>

O

> N > C

3 F

<

N

<

O

< C

4

F > N > O < C

Explanation:

On moving from left to right in a $2^{nd}$ period electronegativity increases.
$\begin{array}{lllll} Element & C & N & O & F \\ EN, Value & 2.5 & 3.0 & 3.5 & 4 \end{array}$
Hence, the order of $EN$ are-
$F > O > N > C$

CG PET - 2018

Classification of Elements and Periodicity in Properties

89519 Pauling's equation for determining the electronegativity of an element is

1

X_{A} - X_{B} = 0.208 \sqrt{Δ}

2

X_{A} + X_{B} = 0.208 \sqrt{Δ}

3

X_{A} - X_{B} = 0.208 Δ^{2}

4

X_{A} - X_{B} = \sqrt{Δ}

Explanation:

According to Pauiling equation.
$\begin{aligned} X_{A} - X_{B} = 0.208 {[E_{A - B} - {(E_{A - A} \times E_{B - B})}^{1 / 2}]}^{1 / 2} \\ X_{A} - X_{B} = 0.208 \sqrt{Δ} \end{aligned}$
Where, $Δ$ is the difference of bond dissociation energies of $A - A$ and B-B bonds.
$X_{A}$ and $X_{B}$ are the electronegativities of $A$ and $B$ respectively.
The factor 0.208 arises from the conversion of Kcals to electron volt.

J and K CET-(2006)

Classification of Elements and Periodicity in Properties

89526 The electronegativity of the given elements increases in the order

1 C, N, Si, P

2 N, Si, C, P

3

Si, P, C, N

4 P, Si, N, C

Explanation:

Electro negativity of given element is-
$\begin{aligned} Si = 1.9 \\ P = 2.19 \\ C = 2.55 \\ N = 3.04 \end{aligned}$
Therefore the increasing order is $- Si < P < C < N$ .

AP EAPCET 20.08.2021 Shift-II

Classification of Elements and Periodicity in Properties

89528 Which pair of elements has maximum electro negativity difference?

1 Li \& F

2

Na & F

3

Na & Br

4

Na & Cl

Explanation:

Among the following $Li$ and $Na$ are the most electropositive and $F, Cl$ and $Br$ are most electronegative.
So, the electronegative of the species are-
| Atom | $F$ | $C l$ | $B r$ | $I$ | $L i$ | $N a$ |
|---|---|---|---|---|---|---|
|EN Value | 4 | 3 | 2.8 | 2.5 | 1 | 0.9 |
|
The difference in electronegativity are.
(i) $F - Li = 4 - 2.5 = 1.5$
(ii) $F - Na = 4 - 0.9 = 3.1$
(iii) $Br - Na = 2.8 - 0.9 = 1.9$
(iv) $Cl - Na = 3 - 0.9 = 2.1$
Hence, the $Na$ and $F$ pairs of elements has maximum electronegativity difference.

AIIMS 26 May 2019 (Evening)

Classification of Elements and Periodicity in Properties

89521 Which of the following is not true for oxidation?

1 Addition of oxygen

2 addition of electronegative element

3 removal of hydrogen

4 removal of electronegative element

Kerala-CEE-29.08.2021

Classification of Elements and Periodicity in Properties

89518 Which of the following indicates the correct variation in electronegativities?

1 F

>

N

<

O

> C

2 F

>

O

> N > C

3 F

<

N

<

O

< C

4

F > N > O < C

Explanation:

On moving from left to right in a $2^{nd}$ period electronegativity increases.
$\begin{array}{lllll} Element & C & N & O & F \\ EN, Value & 2.5 & 3.0 & 3.5 & 4 \end{array}$
Hence, the order of $EN$ are-
$F > O > N > C$

CG PET - 2018

Classification of Elements and Periodicity in Properties

89519 Pauling's equation for determining the electronegativity of an element is

1

X_{A} - X_{B} = 0.208 \sqrt{Δ}

2

X_{A} + X_{B} = 0.208 \sqrt{Δ}

3

X_{A} - X_{B} = 0.208 Δ^{2}

4

X_{A} - X_{B} = \sqrt{Δ}

Explanation:

According to Pauiling equation.
$\begin{aligned} X_{A} - X_{B} = 0.208 {[E_{A - B} - {(E_{A - A} \times E_{B - B})}^{1 / 2}]}^{1 / 2} \\ X_{A} - X_{B} = 0.208 \sqrt{Δ} \end{aligned}$
Where, $Δ$ is the difference of bond dissociation energies of $A - A$ and B-B bonds.
$X_{A}$ and $X_{B}$ are the electronegativities of $A$ and $B$ respectively.
The factor 0.208 arises from the conversion of Kcals to electron volt.

J and K CET-(2006)

Classification of Elements and Periodicity in Properties

89526 The electronegativity of the given elements increases in the order

1 C, N, Si, P

2 N, Si, C, P

3

Si, P, C, N

4 P, Si, N, C

Explanation:

Electro negativity of given element is-
$\begin{aligned} Si = 1.9 \\ P = 2.19 \\ C = 2.55 \\ N = 3.04 \end{aligned}$
Therefore the increasing order is $- Si < P < C < N$ .

AP EAPCET 20.08.2021 Shift-II

Classification of Elements and Periodicity in Properties

89528 Which pair of elements has maximum electro negativity difference?

1 Li \& F

2

Na & F

3

Na & Br

4

Na & Cl

Explanation:

Among the following $Li$ and $Na$ are the most electropositive and $F, Cl$ and $Br$ are most electronegative.
So, the electronegative of the species are-
| Atom | $F$ | $C l$ | $B r$ | $I$ | $L i$ | $N a$ |
|---|---|---|---|---|---|---|
|EN Value | 4 | 3 | 2.8 | 2.5 | 1 | 0.9 |
|
The difference in electronegativity are.
(i) $F - Li = 4 - 2.5 = 1.5$
(ii) $F - Na = 4 - 0.9 = 3.1$
(iii) $Br - Na = 2.8 - 0.9 = 1.9$
(iv) $Cl - Na = 3 - 0.9 = 2.1$
Hence, the $Na$ and $F$ pairs of elements has maximum electronegativity difference.

AIIMS 26 May 2019 (Evening)