QBManager

Classification of Elements and Periodicity in Properties

89516 The electronegativity of the following 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:

$Si$ and $P$ are in the $3^{rd}$ period while $C$ and $N$ are in the $2^{nd}$ period. Elements in $2^{nd}$ period have higher electronegativities than those in the $3^{rd}$ period. Since $N$ has smaller size and has higher nuclear charge than $C$ , its electronegativity is higher than that of C. Similarly, the electronegativity of $P$ is higher than that of $Si$ . Thus, the overall order is : Si, P, C, N.

Karnataka CET-2017

Classification of Elements and Periodicity in Properties

89517 Pauling's electronegativity values for elements are useful in predicting

1 polarity of the molecules

2 position in the EMF series

3 coordination numbers

4 dipole moments

BCECE-2011

Classification of Elements and Periodicity in Properties

89544 Which one of the following represents the correct order of electronegativity?

1

P > O > N

2

N > P > O

3 O

> N > P

4

N > O > P

Explanation:

Electro-negativity is defined as the ability of an atom to attract shared electrons in a covalent bond. The correct order of electro-negativity is $O > N > P$ .

A.P.EAMCET 2002

Classification of Elements and Periodicity in Properties

89524 ${HClO}_{4}$ is a stronger acid than $H_{2} {SO}_{4}$ due to which of the following reasons?
(i) Oxidation state of $Cl$ in ${HClO}_{4}$ is +7 and that of $S$ in $H_{2} {SO}_{4}$ is +6
(ii) $Cl$ is more electronegative then $S$
(iii) ${ClO}_{3}$ part of ${HClO}_{4}$ can break $O$ -H bond more easily to liberate a proton than ${SO}_{3}$ is dibasic
(iv) ${ClO}_{3}$ part of ${HClO}_{4}$ can break $O$ -H bond more easily to liberate a proton than ${SO}_{3}$ part in $H_{2} {SO}_{4}$

1 (i), (ii) and (iii) only

2 (i), (ii), and (iv) only

3 (i), (iii), and (iv) only

4 (ii), (iii), (iv) only

Explanation:

In ${HClO}_{4}$ Oxidation state of $Cl$ is +7 and in $H_{2} {SO}_{4}$ Oxidation state of $S$ is $+ 6, Cl$ is more electronegative than $S$ . ${HClO}_{4}$ is mono acidic where as $H_{2} {SO}_{4}$ is dibasic ${ClO}_{3}$ part of ${HClO}_{4}$ can break $O - H$ bond more easily to liberate a proton than ${SO}_{3}$ part in $H_{2} {SO}_{4}$ . So, the option (b) is correct.

AP EAPCET 24.08.2021 Shift-II

Classification of Elements and Periodicity in Properties

89516 The electronegativity of the following 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:

$Si$ and $P$ are in the $3^{rd}$ period while $C$ and $N$ are in the $2^{nd}$ period. Elements in $2^{nd}$ period have higher electronegativities than those in the $3^{rd}$ period. Since $N$ has smaller size and has higher nuclear charge than $C$ , its electronegativity is higher than that of C. Similarly, the electronegativity of $P$ is higher than that of $Si$ . Thus, the overall order is : Si, P, C, N.

Karnataka CET-2017

Classification of Elements and Periodicity in Properties

89517 Pauling's electronegativity values for elements are useful in predicting

1 polarity of the molecules

2 position in the EMF series

3 coordination numbers

4 dipole moments

BCECE-2011

Classification of Elements and Periodicity in Properties

89544 Which one of the following represents the correct order of electronegativity?

1

P > O > N

2

N > P > O

3 O

> N > P

4

N > O > P

Explanation:

Electro-negativity is defined as the ability of an atom to attract shared electrons in a covalent bond. The correct order of electro-negativity is $O > N > P$ .

A.P.EAMCET 2002

Classification of Elements and Periodicity in Properties

89524 ${HClO}_{4}$ is a stronger acid than $H_{2} {SO}_{4}$ due to which of the following reasons?
(i) Oxidation state of $Cl$ in ${HClO}_{4}$ is +7 and that of $S$ in $H_{2} {SO}_{4}$ is +6
(ii) $Cl$ is more electronegative then $S$
(iii) ${ClO}_{3}$ part of ${HClO}_{4}$ can break $O$ -H bond more easily to liberate a proton than ${SO}_{3}$ is dibasic
(iv) ${ClO}_{3}$ part of ${HClO}_{4}$ can break $O$ -H bond more easily to liberate a proton than ${SO}_{3}$ part in $H_{2} {SO}_{4}$

1 (i), (ii) and (iii) only

2 (i), (ii), and (iv) only

3 (i), (iii), and (iv) only

4 (ii), (iii), (iv) only

Explanation:

In ${HClO}_{4}$ Oxidation state of $Cl$ is +7 and in $H_{2} {SO}_{4}$ Oxidation state of $S$ is $+ 6, Cl$ is more electronegative than $S$ . ${HClO}_{4}$ is mono acidic where as $H_{2} {SO}_{4}$ is dibasic ${ClO}_{3}$ part of ${HClO}_{4}$ can break $O - H$ bond more easily to liberate a proton than ${SO}_{3}$ part in $H_{2} {SO}_{4}$ . So, the option (b) is correct.

AP EAPCET 24.08.2021 Shift-II

Classification of Elements and Periodicity in Properties

89516 The electronegativity of the following 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:

$Si$ and $P$ are in the $3^{rd}$ period while $C$ and $N$ are in the $2^{nd}$ period. Elements in $2^{nd}$ period have higher electronegativities than those in the $3^{rd}$ period. Since $N$ has smaller size and has higher nuclear charge than $C$ , its electronegativity is higher than that of C. Similarly, the electronegativity of $P$ is higher than that of $Si$ . Thus, the overall order is : Si, P, C, N.

Karnataka CET-2017

Classification of Elements and Periodicity in Properties

89517 Pauling's electronegativity values for elements are useful in predicting

1 polarity of the molecules

2 position in the EMF series

3 coordination numbers

4 dipole moments

BCECE-2011

Classification of Elements and Periodicity in Properties

89544 Which one of the following represents the correct order of electronegativity?

1

P > O > N

2

N > P > O

3 O

> N > P

4

N > O > P

Explanation:

Electro-negativity is defined as the ability of an atom to attract shared electrons in a covalent bond. The correct order of electro-negativity is $O > N > P$ .

A.P.EAMCET 2002

Classification of Elements and Periodicity in Properties

89524 ${HClO}_{4}$ is a stronger acid than $H_{2} {SO}_{4}$ due to which of the following reasons?
(i) Oxidation state of $Cl$ in ${HClO}_{4}$ is +7 and that of $S$ in $H_{2} {SO}_{4}$ is +6
(ii) $Cl$ is more electronegative then $S$
(iii) ${ClO}_{3}$ part of ${HClO}_{4}$ can break $O$ -H bond more easily to liberate a proton than ${SO}_{3}$ is dibasic
(iv) ${ClO}_{3}$ part of ${HClO}_{4}$ can break $O$ -H bond more easily to liberate a proton than ${SO}_{3}$ part in $H_{2} {SO}_{4}$

1 (i), (ii) and (iii) only

2 (i), (ii), and (iv) only

3 (i), (iii), and (iv) only

4 (ii), (iii), (iv) only

Explanation:

In ${HClO}_{4}$ Oxidation state of $Cl$ is +7 and in $H_{2} {SO}_{4}$ Oxidation state of $S$ is $+ 6, Cl$ is more electronegative than $S$ . ${HClO}_{4}$ is mono acidic where as $H_{2} {SO}_{4}$ is dibasic ${ClO}_{3}$ part of ${HClO}_{4}$ can break $O - H$ bond more easily to liberate a proton than ${SO}_{3}$ part in $H_{2} {SO}_{4}$ . So, the option (b) is correct.

AP EAPCET 24.08.2021 Shift-II

Classification of Elements and Periodicity in Properties

89516 The electronegativity of the following 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:

$Si$ and $P$ are in the $3^{rd}$ period while $C$ and $N$ are in the $2^{nd}$ period. Elements in $2^{nd}$ period have higher electronegativities than those in the $3^{rd}$ period. Since $N$ has smaller size and has higher nuclear charge than $C$ , its electronegativity is higher than that of C. Similarly, the electronegativity of $P$ is higher than that of $Si$ . Thus, the overall order is : Si, P, C, N.

Karnataka CET-2017

Classification of Elements and Periodicity in Properties

89517 Pauling's electronegativity values for elements are useful in predicting

1 polarity of the molecules

2 position in the EMF series

3 coordination numbers

4 dipole moments

BCECE-2011

Classification of Elements and Periodicity in Properties

89544 Which one of the following represents the correct order of electronegativity?

1

P > O > N

2

N > P > O

3 O

> N > P

4

N > O > P

Explanation:

Electro-negativity is defined as the ability of an atom to attract shared electrons in a covalent bond. The correct order of electro-negativity is $O > N > P$ .

A.P.EAMCET 2002

Classification of Elements and Periodicity in Properties

89524 ${HClO}_{4}$ is a stronger acid than $H_{2} {SO}_{4}$ due to which of the following reasons?
(i) Oxidation state of $Cl$ in ${HClO}_{4}$ is +7 and that of $S$ in $H_{2} {SO}_{4}$ is +6
(ii) $Cl$ is more electronegative then $S$
(iii) ${ClO}_{3}$ part of ${HClO}_{4}$ can break $O$ -H bond more easily to liberate a proton than ${SO}_{3}$ is dibasic
(iv) ${ClO}_{3}$ part of ${HClO}_{4}$ can break $O$ -H bond more easily to liberate a proton than ${SO}_{3}$ part in $H_{2} {SO}_{4}$

1 (i), (ii) and (iii) only

2 (i), (ii), and (iv) only

3 (i), (iii), and (iv) only

4 (ii), (iii), (iv) only

Explanation:

In ${HClO}_{4}$ Oxidation state of $Cl$ is +7 and in $H_{2} {SO}_{4}$ Oxidation state of $S$ is $+ 6, Cl$ is more electronegative than $S$ . ${HClO}_{4}$ is mono acidic where as $H_{2} {SO}_{4}$ is dibasic ${ClO}_{3}$ part of ${HClO}_{4}$ can break $O - H$ bond more easily to liberate a proton than ${SO}_{3}$ part in $H_{2} {SO}_{4}$ . So, the option (b) is correct.

AP EAPCET 24.08.2021 Shift-II