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Structure of Atom

238610 The group having isoelectronic species is
#[Qdiff: Easy, QCat: Theory Based, examname: |---|---|, |
,
, {|c|c|c|}
, \mathrm{F}^{-}, (JEE Main-2017)
, | Species | $l A t o m i c |, | n u m b e r$ | $l N u m b e r o f |, | e l e c t r o n s$ |, | $^{2 -}$ | 8 | 10 |, | $^{-}$ | 9 | 10 |, | $^{+}$ | 11 | 10 |, | $^{2 +}$ | 12 | 10 |, | $^{-}$ | 8 | 9 |, |$$ | 11 | 11 |, | $^{+}$ | 12 | 11 |, Option (a) is correct which contains isoelectronic species $O^{2 -}, {Na}^{+}, {Mg}^{2 +}$
, 75. Which one of the following sets of ions represents a collection of isoelectronic species?
,

1

O^{2 -}, F^{-}, {Na}^{+}, {Mg}^{2 +}

2

O^{-}, F^{-}, Na, {Mg}^{+}

3

O^{2 -}, F^{-}, Na, {Mg}^{2 +}

4

O^{-}, F^{-}, {Na}^{+}, {Mg}^{2 +}

Explanation:

The ions which have the same number of electrons are called isoelectronic
$\begin{array}{cccc} K^{+} & {Cl}^{-} & {Ca}^{2 +} & {Sc}^{3 +} \\ (2, 8, 8) & (2, 8, 8) & (2, 8, 8) & (2, 8, 8) \end{array}$
All these ions contains 18 electrons each. So all these ions are isoelectronic.

Structure of Atom

238618 Negatively charged particles are called
#[Qdiff: Easy, QCat: Theory Based, examname: |---|, Exp: } |, |
, {|c|c|c|}
, J and K CET-(2014)
, | { Ans. A
|, |Particles | Symbol | Relative charge |, |Electrons | $| - 1 |, | P r o t o n s |$ | +1 |, |Neutrons | $$ | 0 |, 85. Mass number of an atom is the sum of
,

1 Electrons

2 Protons

3 Neutrons

4 None of the above

Explanation:

} |, |
, {|c|c|c|}
, J and K CET-(2014)
, | { Ans. A
|, |Particles | Symbol | Relative charge |, |Electrons | $| - 1 |, | P r o t o n s |$ | +1 |, |Neutrons | $$ | 0 |, 85. Mass number of an atom is the sum of
, (a.) Number of protons + number of neutrons + number of electrons
, (b.) Number of protons + number of neutrons
, (c.) Number of protons + number of electrons
, (d.) Number of electrons + number of neutrons.]#
(a.) Electrons
(b.) Protons
(c.) Neutrons
(d.) None of the above
Ans: b
Exp: The total number of nucleous is termed as mass number (A) of the atom.
Hence, mass number $(A) =$ No. of protons $(p) +$ No. of neutrons(n).

Structure of Atom

238622 The specific heat of a metal is 0.11 and its equivalent weight is 18.61. Its exact atomic weight is

1 58.2

2 29.1

3 55.83

4 27.91

Explanation:

Approximate atomic wt. of the metal
$\begin{matrix} = \frac{6.4}{specificheat} = \frac{6.4}{0.11} = 58.18 \\ Valency of the metal = \frac{Approx at.wt}{Eq.wt.} \\ = \frac{58.18}{18.61} = 3 (wholeno.) \end{matrix}$
So, exact at. wt. of the element $=$ eq.wt. $\times$ Valency
$\begin{aligned} = 18.61 \times 3 \\ = 55.83 \end{aligned}$

J and K CET-(1998)

Structure of Atom

238627 An element with mass number 81 contains $31.7 %$ more neutrons as compared to protons. It symbol is

1

_{40} X^{81}

2

_{35} X^{81}

3

_{81} X^{317}

4

_{40} X^{81}

Explanation:

Let, the number of protons $= x$
Number of neutrons $= x + \frac{31.7 x}{100} = \frac{131.7 x}{100}$
Mass number $=$ number of protons + number of neutrons $81 = x + \frac{131.7 x}{100}$
$∴ x = 35$
Thus, the symbol of the element is $_{35} X^{81}$ .

JIPMER-2019

Structure of Atom

238610 The group having isoelectronic species is
#[Qdiff: Easy, QCat: Theory Based, examname: |---|---|, |
,
, {|c|c|c|}
, \mathrm{F}^{-}, (JEE Main-2017)
, | Species | $l A t o m i c |, | n u m b e r$ | $l N u m b e r o f |, | e l e c t r o n s$ |, | $^{2 -}$ | 8 | 10 |, | $^{-}$ | 9 | 10 |, | $^{+}$ | 11 | 10 |, | $^{2 +}$ | 12 | 10 |, | $^{-}$ | 8 | 9 |, |$$ | 11 | 11 |, | $^{+}$ | 12 | 11 |, Option (a) is correct which contains isoelectronic species $O^{2 -}, {Na}^{+}, {Mg}^{2 +}$
, 75. Which one of the following sets of ions represents a collection of isoelectronic species?
,

1

O^{2 -}, F^{-}, {Na}^{+}, {Mg}^{2 +}

2

O^{-}, F^{-}, Na, {Mg}^{+}

3

O^{2 -}, F^{-}, Na, {Mg}^{2 +}

4

O^{-}, F^{-}, {Na}^{+}, {Mg}^{2 +}

Explanation:

The ions which have the same number of electrons are called isoelectronic
$\begin{array}{cccc} K^{+} & {Cl}^{-} & {Ca}^{2 +} & {Sc}^{3 +} \\ (2, 8, 8) & (2, 8, 8) & (2, 8, 8) & (2, 8, 8) \end{array}$
All these ions contains 18 electrons each. So all these ions are isoelectronic.

Structure of Atom

238618 Negatively charged particles are called
#[Qdiff: Easy, QCat: Theory Based, examname: |---|, Exp: } |, |
, {|c|c|c|}
, J and K CET-(2014)
, | { Ans. A
|, |Particles | Symbol | Relative charge |, |Electrons | $| - 1 |, | P r o t o n s |$ | +1 |, |Neutrons | $$ | 0 |, 85. Mass number of an atom is the sum of
,

1 Electrons

2 Protons

3 Neutrons

4 None of the above

Explanation:

} |, |
, {|c|c|c|}
, J and K CET-(2014)
, | { Ans. A
|, |Particles | Symbol | Relative charge |, |Electrons | $| - 1 |, | P r o t o n s |$ | +1 |, |Neutrons | $$ | 0 |, 85. Mass number of an atom is the sum of
, (a.) Number of protons + number of neutrons + number of electrons
, (b.) Number of protons + number of neutrons
, (c.) Number of protons + number of electrons
, (d.) Number of electrons + number of neutrons.]#
(a.) Electrons
(b.) Protons
(c.) Neutrons
(d.) None of the above
Ans: b
Exp: The total number of nucleous is termed as mass number (A) of the atom.
Hence, mass number $(A) =$ No. of protons $(p) +$ No. of neutrons(n).

Structure of Atom

238622 The specific heat of a metal is 0.11 and its equivalent weight is 18.61. Its exact atomic weight is

1 58.2

2 29.1

3 55.83

4 27.91

Explanation:

Approximate atomic wt. of the metal
$\begin{matrix} = \frac{6.4}{specificheat} = \frac{6.4}{0.11} = 58.18 \\ Valency of the metal = \frac{Approx at.wt}{Eq.wt.} \\ = \frac{58.18}{18.61} = 3 (wholeno.) \end{matrix}$
So, exact at. wt. of the element $=$ eq.wt. $\times$ Valency
$\begin{aligned} = 18.61 \times 3 \\ = 55.83 \end{aligned}$

J and K CET-(1998)

Structure of Atom

238627 An element with mass number 81 contains $31.7 %$ more neutrons as compared to protons. It symbol is

1

_{40} X^{81}

2

_{35} X^{81}

3

_{81} X^{317}

4

_{40} X^{81}

Explanation:

Let, the number of protons $= x$
Number of neutrons $= x + \frac{31.7 x}{100} = \frac{131.7 x}{100}$
Mass number $=$ number of protons + number of neutrons $81 = x + \frac{131.7 x}{100}$
$∴ x = 35$
Thus, the symbol of the element is $_{35} X^{81}$ .

JIPMER-2019

Structure of Atom

238610 The group having isoelectronic species is
#[Qdiff: Easy, QCat: Theory Based, examname: |---|---|, |
,
, {|c|c|c|}
, \mathrm{F}^{-}, (JEE Main-2017)
, | Species | $l A t o m i c |, | n u m b e r$ | $l N u m b e r o f |, | e l e c t r o n s$ |, | $^{2 -}$ | 8 | 10 |, | $^{-}$ | 9 | 10 |, | $^{+}$ | 11 | 10 |, | $^{2 +}$ | 12 | 10 |, | $^{-}$ | 8 | 9 |, |$$ | 11 | 11 |, | $^{+}$ | 12 | 11 |, Option (a) is correct which contains isoelectronic species $O^{2 -}, {Na}^{+}, {Mg}^{2 +}$
, 75. Which one of the following sets of ions represents a collection of isoelectronic species?
,

1

O^{2 -}, F^{-}, {Na}^{+}, {Mg}^{2 +}

2

O^{-}, F^{-}, Na, {Mg}^{+}

3

O^{2 -}, F^{-}, Na, {Mg}^{2 +}

4

O^{-}, F^{-}, {Na}^{+}, {Mg}^{2 +}

Explanation:

The ions which have the same number of electrons are called isoelectronic
$\begin{array}{cccc} K^{+} & {Cl}^{-} & {Ca}^{2 +} & {Sc}^{3 +} \\ (2, 8, 8) & (2, 8, 8) & (2, 8, 8) & (2, 8, 8) \end{array}$
All these ions contains 18 electrons each. So all these ions are isoelectronic.

Structure of Atom

238618 Negatively charged particles are called
#[Qdiff: Easy, QCat: Theory Based, examname: |---|, Exp: } |, |
, {|c|c|c|}
, J and K CET-(2014)
, | { Ans. A
|, |Particles | Symbol | Relative charge |, |Electrons | $| - 1 |, | P r o t o n s |$ | +1 |, |Neutrons | $$ | 0 |, 85. Mass number of an atom is the sum of
,

1 Electrons

2 Protons

3 Neutrons

4 None of the above

Explanation:

} |, |
, {|c|c|c|}
, J and K CET-(2014)
, | { Ans. A
|, |Particles | Symbol | Relative charge |, |Electrons | $| - 1 |, | P r o t o n s |$ | +1 |, |Neutrons | $$ | 0 |, 85. Mass number of an atom is the sum of
, (a.) Number of protons + number of neutrons + number of electrons
, (b.) Number of protons + number of neutrons
, (c.) Number of protons + number of electrons
, (d.) Number of electrons + number of neutrons.]#
(a.) Electrons
(b.) Protons
(c.) Neutrons
(d.) None of the above
Ans: b
Exp: The total number of nucleous is termed as mass number (A) of the atom.
Hence, mass number $(A) =$ No. of protons $(p) +$ No. of neutrons(n).

Structure of Atom

238622 The specific heat of a metal is 0.11 and its equivalent weight is 18.61. Its exact atomic weight is

1 58.2

2 29.1

3 55.83

4 27.91

Explanation:

Approximate atomic wt. of the metal
$\begin{matrix} = \frac{6.4}{specificheat} = \frac{6.4}{0.11} = 58.18 \\ Valency of the metal = \frac{Approx at.wt}{Eq.wt.} \\ = \frac{58.18}{18.61} = 3 (wholeno.) \end{matrix}$
So, exact at. wt. of the element $=$ eq.wt. $\times$ Valency
$\begin{aligned} = 18.61 \times 3 \\ = 55.83 \end{aligned}$

J and K CET-(1998)

Structure of Atom

238627 An element with mass number 81 contains $31.7 %$ more neutrons as compared to protons. It symbol is

1

_{40} X^{81}

2

_{35} X^{81}

3

_{81} X^{317}

4

_{40} X^{81}

Explanation:

Let, the number of protons $= x$
Number of neutrons $= x + \frac{31.7 x}{100} = \frac{131.7 x}{100}$
Mass number $=$ number of protons + number of neutrons $81 = x + \frac{131.7 x}{100}$
$∴ x = 35$
Thus, the symbol of the element is $_{35} X^{81}$ .

JIPMER-2019

Structure of Atom

238610 The group having isoelectronic species is
#[Qdiff: Easy, QCat: Theory Based, examname: |---|---|, |
,
, {|c|c|c|}
, \mathrm{F}^{-}, (JEE Main-2017)
, | Species | $l A t o m i c |, | n u m b e r$ | $l N u m b e r o f |, | e l e c t r o n s$ |, | $^{2 -}$ | 8 | 10 |, | $^{-}$ | 9 | 10 |, | $^{+}$ | 11 | 10 |, | $^{2 +}$ | 12 | 10 |, | $^{-}$ | 8 | 9 |, |$$ | 11 | 11 |, | $^{+}$ | 12 | 11 |, Option (a) is correct which contains isoelectronic species $O^{2 -}, {Na}^{+}, {Mg}^{2 +}$
, 75. Which one of the following sets of ions represents a collection of isoelectronic species?
,

1

O^{2 -}, F^{-}, {Na}^{+}, {Mg}^{2 +}

2

O^{-}, F^{-}, Na, {Mg}^{+}

3

O^{2 -}, F^{-}, Na, {Mg}^{2 +}

4

O^{-}, F^{-}, {Na}^{+}, {Mg}^{2 +}

Explanation:

The ions which have the same number of electrons are called isoelectronic
$\begin{array}{cccc} K^{+} & {Cl}^{-} & {Ca}^{2 +} & {Sc}^{3 +} \\ (2, 8, 8) & (2, 8, 8) & (2, 8, 8) & (2, 8, 8) \end{array}$
All these ions contains 18 electrons each. So all these ions are isoelectronic.

Structure of Atom

238618 Negatively charged particles are called
#[Qdiff: Easy, QCat: Theory Based, examname: |---|, Exp: } |, |
, {|c|c|c|}
, J and K CET-(2014)
, | { Ans. A
|, |Particles | Symbol | Relative charge |, |Electrons | $| - 1 |, | P r o t o n s |$ | +1 |, |Neutrons | $$ | 0 |, 85. Mass number of an atom is the sum of
,

1 Electrons

2 Protons

3 Neutrons

4 None of the above

Explanation:

} |, |
, {|c|c|c|}
, J and K CET-(2014)
, | { Ans. A
|, |Particles | Symbol | Relative charge |, |Electrons | $| - 1 |, | P r o t o n s |$ | +1 |, |Neutrons | $$ | 0 |, 85. Mass number of an atom is the sum of
, (a.) Number of protons + number of neutrons + number of electrons
, (b.) Number of protons + number of neutrons
, (c.) Number of protons + number of electrons
, (d.) Number of electrons + number of neutrons.]#
(a.) Electrons
(b.) Protons
(c.) Neutrons
(d.) None of the above
Ans: b
Exp: The total number of nucleous is termed as mass number (A) of the atom.
Hence, mass number $(A) =$ No. of protons $(p) +$ No. of neutrons(n).

Structure of Atom

238622 The specific heat of a metal is 0.11 and its equivalent weight is 18.61. Its exact atomic weight is

1 58.2

2 29.1

3 55.83

4 27.91

Explanation:

Approximate atomic wt. of the metal
$\begin{matrix} = \frac{6.4}{specificheat} = \frac{6.4}{0.11} = 58.18 \\ Valency of the metal = \frac{Approx at.wt}{Eq.wt.} \\ = \frac{58.18}{18.61} = 3 (wholeno.) \end{matrix}$
So, exact at. wt. of the element $=$ eq.wt. $\times$ Valency
$\begin{aligned} = 18.61 \times 3 \\ = 55.83 \end{aligned}$

J and K CET-(1998)

Structure of Atom

238627 An element with mass number 81 contains $31.7 %$ more neutrons as compared to protons. It symbol is

1

_{40} X^{81}

2

_{35} X^{81}

3

_{81} X^{317}

4

_{40} X^{81}

Explanation:

Let, the number of protons $= x$
Number of neutrons $= x + \frac{31.7 x}{100} = \frac{131.7 x}{100}$
Mass number $=$ number of protons + number of neutrons $81 = x + \frac{131.7 x}{100}$
$∴ x = 35$
Thus, the symbol of the element is $_{35} X^{81}$ .

JIPMER-2019

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