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CHXII08:THE D- & F-BLOCK ELEMENTS

321487 Which of the following will show increase in weight when kept in magnetic field?

1

{TiO}_{2}

2

{Fe}_{2} {({SO}_{4})}_{3}

3

{KMnO}_{4}

4

{ScCl}_{3}

Explanation:

${Fe}^{+ 3} - 3 d^{5}$ (5 unpaired electrons).
Fe will be attracted in magnetic field so it shows increase in weight.

CHXII08:THE D- & F-BLOCK ELEMENTS

321488 The highest magnetic moment is shown by the transition metal ion with the configuration

1

3 d^{2}

2

3 d^{5}

3

3 d^{7}

4

3 d^{9}

Explanation:

Magnetic moment $= \sqrt{n (n + 2)} B M$
where $n =$ number of unpaired electrons.
Since $3 d^{5}$ has maximum number of unpaired electrons, its magnetic moment will be highest.

KCET - 2006

CHXII08:THE D- & F-BLOCK ELEMENTS

321489 Among the following, the incorrect statement is

1 The fourth period atom having a total of six '

d

' electrons is

Fe

2

{Ca}^{2 +}

has a smaller radius than

K^{+}

, because it has a higher nuclear charge

3 The magnetic moment for

{Co}^{3 +}

is

4 .90 B M

4 Among

{Fe}^{2 +}

and

{Fe}^{3 +}

, the smaller ion is less paramagnetic.

Explanation:

${Fe}^{3 +}$ is smaller in size. Its electronic configuration is $[A r] 3 d^{5}$ . It has 5 unpaired electrons and thus is more paramagnetic than ${Fe}^{2 +}$ ion (which has only 4 unpaired electrons)

CHXII08:THE D- & F-BLOCK ELEMENTS

321490 For which of the following pairs, magnetic moment is same?

1

{MnCl}_{2}, {CuSO}_{4}

2

{CuCl}_{2}, {TiCl}_{3}

3

{TiO}_{2}, {CuSO}_{4}

4

{TiCl}_{3}, {NiCl}_{2}

Explanation:

Species having the same number of unpaired electrons, have same magnetic moment.
(1) ${MnCl}_{2} \Rightarrow {Mn}^{2 +} = [Ar] 3 d^{5}, 4 s^{0}$
(five unpaired electrons)
${CuSO}_{4} \Rightarrow {Cu}^{2 +} = [Ar] 3 d^{9}, 4 s^{0}$
(one unpaired electron)
(2) ${CuCl}_{2} \Rightarrow {Cu}^{2 +} = [Ar] 3 d^{9}, 4 s^{0}$
(one unpaired electron)
${TiCl}_{3} \Rightarrow {Ti}^{3 +} = [Ar] 3 d^{1}, 4 s^{0}$
(one unpaired electron)
(3) ${TiO}_{2} \Rightarrow {Ti}^{4 +} = [Ar] 3 d^{0}, 4 s^{0}$
(no unpaired electron)
${CuSO}_{4} \Rightarrow {Cu}^{2 +} = [Ar] 3 d^{9}, 4 s^{0}$
(one unpaired electron)
(4) ${TiCl}_{3} \Rightarrow {Ti}^{3 +} = [Ar] 3 d^{1}, 4 s^{0}$
(one unpaired electron)
${NiCl}_{2} \Rightarrow {Ni}^{2 +} = [Ar] 3 d^{8}, 4 s^{0}$
(two unpaired electrons)
$∴ {CuCl}_{2}$ and ${TiCl}_{3}$ will have same magnetic moment.
$∴ μ = \sqrt{1 (1 + 2)} = \sqrt{3} = 1.73 BM$

AIIMS - 2009

CHXII08:THE D- & F-BLOCK ELEMENTS

321487 Which of the following will show increase in weight when kept in magnetic field?

1

{TiO}_{2}

2

{Fe}_{2} {({SO}_{4})}_{3}

3

{KMnO}_{4}

4

{ScCl}_{3}

Explanation:

${Fe}^{+ 3} - 3 d^{5}$ (5 unpaired electrons).
Fe will be attracted in magnetic field so it shows increase in weight.

CHXII08:THE D- & F-BLOCK ELEMENTS

321488 The highest magnetic moment is shown by the transition metal ion with the configuration

1

3 d^{2}

2

3 d^{5}

3

3 d^{7}

4

3 d^{9}

Explanation:

Magnetic moment $= \sqrt{n (n + 2)} B M$
where $n =$ number of unpaired electrons.
Since $3 d^{5}$ has maximum number of unpaired electrons, its magnetic moment will be highest.

KCET - 2006

CHXII08:THE D- & F-BLOCK ELEMENTS

321489 Among the following, the incorrect statement is

1 The fourth period atom having a total of six '

d

' electrons is

Fe

2

{Ca}^{2 +}

has a smaller radius than

K^{+}

, because it has a higher nuclear charge

3 The magnetic moment for

{Co}^{3 +}

is

4 .90 B M

4 Among

{Fe}^{2 +}

and

{Fe}^{3 +}

, the smaller ion is less paramagnetic.

Explanation:

${Fe}^{3 +}$ is smaller in size. Its electronic configuration is $[A r] 3 d^{5}$ . It has 5 unpaired electrons and thus is more paramagnetic than ${Fe}^{2 +}$ ion (which has only 4 unpaired electrons)

CHXII08:THE D- & F-BLOCK ELEMENTS

321490 For which of the following pairs, magnetic moment is same?

1

{MnCl}_{2}, {CuSO}_{4}

2

{CuCl}_{2}, {TiCl}_{3}

3

{TiO}_{2}, {CuSO}_{4}

4

{TiCl}_{3}, {NiCl}_{2}

Explanation:

Species having the same number of unpaired electrons, have same magnetic moment.
(1) ${MnCl}_{2} \Rightarrow {Mn}^{2 +} = [Ar] 3 d^{5}, 4 s^{0}$
(five unpaired electrons)
${CuSO}_{4} \Rightarrow {Cu}^{2 +} = [Ar] 3 d^{9}, 4 s^{0}$
(one unpaired electron)
(2) ${CuCl}_{2} \Rightarrow {Cu}^{2 +} = [Ar] 3 d^{9}, 4 s^{0}$
(one unpaired electron)
${TiCl}_{3} \Rightarrow {Ti}^{3 +} = [Ar] 3 d^{1}, 4 s^{0}$
(one unpaired electron)
(3) ${TiO}_{2} \Rightarrow {Ti}^{4 +} = [Ar] 3 d^{0}, 4 s^{0}$
(no unpaired electron)
${CuSO}_{4} \Rightarrow {Cu}^{2 +} = [Ar] 3 d^{9}, 4 s^{0}$
(one unpaired electron)
(4) ${TiCl}_{3} \Rightarrow {Ti}^{3 +} = [Ar] 3 d^{1}, 4 s^{0}$
(one unpaired electron)
${NiCl}_{2} \Rightarrow {Ni}^{2 +} = [Ar] 3 d^{8}, 4 s^{0}$
(two unpaired electrons)
$∴ {CuCl}_{2}$ and ${TiCl}_{3}$ will have same magnetic moment.
$∴ μ = \sqrt{1 (1 + 2)} = \sqrt{3} = 1.73 BM$

AIIMS - 2009

CHXII08:THE D- & F-BLOCK ELEMENTS

321487 Which of the following will show increase in weight when kept in magnetic field?

1

{TiO}_{2}

2

{Fe}_{2} {({SO}_{4})}_{3}

3

{KMnO}_{4}

4

{ScCl}_{3}

Explanation:

${Fe}^{+ 3} - 3 d^{5}$ (5 unpaired electrons).
Fe will be attracted in magnetic field so it shows increase in weight.

CHXII08:THE D- & F-BLOCK ELEMENTS

321488 The highest magnetic moment is shown by the transition metal ion with the configuration

1

3 d^{2}

2

3 d^{5}

3

3 d^{7}

4

3 d^{9}

Explanation:

Magnetic moment $= \sqrt{n (n + 2)} B M$
where $n =$ number of unpaired electrons.
Since $3 d^{5}$ has maximum number of unpaired electrons, its magnetic moment will be highest.

KCET - 2006

CHXII08:THE D- & F-BLOCK ELEMENTS

321489 Among the following, the incorrect statement is

1 The fourth period atom having a total of six '

d

' electrons is

Fe

2

{Ca}^{2 +}

has a smaller radius than

K^{+}

, because it has a higher nuclear charge

3 The magnetic moment for

{Co}^{3 +}

is

4 .90 B M

4 Among

{Fe}^{2 +}

and

{Fe}^{3 +}

, the smaller ion is less paramagnetic.

Explanation:

${Fe}^{3 +}$ is smaller in size. Its electronic configuration is $[A r] 3 d^{5}$ . It has 5 unpaired electrons and thus is more paramagnetic than ${Fe}^{2 +}$ ion (which has only 4 unpaired electrons)

CHXII08:THE D- & F-BLOCK ELEMENTS

321490 For which of the following pairs, magnetic moment is same?

1

{MnCl}_{2}, {CuSO}_{4}

2

{CuCl}_{2}, {TiCl}_{3}

3

{TiO}_{2}, {CuSO}_{4}

4

{TiCl}_{3}, {NiCl}_{2}

Explanation:

Species having the same number of unpaired electrons, have same magnetic moment.
(1) ${MnCl}_{2} \Rightarrow {Mn}^{2 +} = [Ar] 3 d^{5}, 4 s^{0}$
(five unpaired electrons)
${CuSO}_{4} \Rightarrow {Cu}^{2 +} = [Ar] 3 d^{9}, 4 s^{0}$
(one unpaired electron)
(2) ${CuCl}_{2} \Rightarrow {Cu}^{2 +} = [Ar] 3 d^{9}, 4 s^{0}$
(one unpaired electron)
${TiCl}_{3} \Rightarrow {Ti}^{3 +} = [Ar] 3 d^{1}, 4 s^{0}$
(one unpaired electron)
(3) ${TiO}_{2} \Rightarrow {Ti}^{4 +} = [Ar] 3 d^{0}, 4 s^{0}$
(no unpaired electron)
${CuSO}_{4} \Rightarrow {Cu}^{2 +} = [Ar] 3 d^{9}, 4 s^{0}$
(one unpaired electron)
(4) ${TiCl}_{3} \Rightarrow {Ti}^{3 +} = [Ar] 3 d^{1}, 4 s^{0}$
(one unpaired electron)
${NiCl}_{2} \Rightarrow {Ni}^{2 +} = [Ar] 3 d^{8}, 4 s^{0}$
(two unpaired electrons)
$∴ {CuCl}_{2}$ and ${TiCl}_{3}$ will have same magnetic moment.
$∴ μ = \sqrt{1 (1 + 2)} = \sqrt{3} = 1.73 BM$

AIIMS - 2009

CHXII08:THE D- & F-BLOCK ELEMENTS

321487 Which of the following will show increase in weight when kept in magnetic field?

1

{TiO}_{2}

2

{Fe}_{2} {({SO}_{4})}_{3}

3

{KMnO}_{4}

4

{ScCl}_{3}

Explanation:

${Fe}^{+ 3} - 3 d^{5}$ (5 unpaired electrons).
Fe will be attracted in magnetic field so it shows increase in weight.

CHXII08:THE D- & F-BLOCK ELEMENTS

321488 The highest magnetic moment is shown by the transition metal ion with the configuration

1

3 d^{2}

2

3 d^{5}

3

3 d^{7}

4

3 d^{9}

Explanation:

Magnetic moment $= \sqrt{n (n + 2)} B M$
where $n =$ number of unpaired electrons.
Since $3 d^{5}$ has maximum number of unpaired electrons, its magnetic moment will be highest.

KCET - 2006

CHXII08:THE D- & F-BLOCK ELEMENTS

321489 Among the following, the incorrect statement is

1 The fourth period atom having a total of six '

d

' electrons is

Fe

2

{Ca}^{2 +}

has a smaller radius than

K^{+}

, because it has a higher nuclear charge

3 The magnetic moment for

{Co}^{3 +}

is

4 .90 B M

4 Among

{Fe}^{2 +}

and

{Fe}^{3 +}

, the smaller ion is less paramagnetic.

Explanation:

${Fe}^{3 +}$ is smaller in size. Its electronic configuration is $[A r] 3 d^{5}$ . It has 5 unpaired electrons and thus is more paramagnetic than ${Fe}^{2 +}$ ion (which has only 4 unpaired electrons)

CHXII08:THE D- & F-BLOCK ELEMENTS

321490 For which of the following pairs, magnetic moment is same?

1

{MnCl}_{2}, {CuSO}_{4}

2

{CuCl}_{2}, {TiCl}_{3}

3

{TiO}_{2}, {CuSO}_{4}

4

{TiCl}_{3}, {NiCl}_{2}

Explanation:

Species having the same number of unpaired electrons, have same magnetic moment.
(1) ${MnCl}_{2} \Rightarrow {Mn}^{2 +} = [Ar] 3 d^{5}, 4 s^{0}$
(five unpaired electrons)
${CuSO}_{4} \Rightarrow {Cu}^{2 +} = [Ar] 3 d^{9}, 4 s^{0}$
(one unpaired electron)
(2) ${CuCl}_{2} \Rightarrow {Cu}^{2 +} = [Ar] 3 d^{9}, 4 s^{0}$
(one unpaired electron)
${TiCl}_{3} \Rightarrow {Ti}^{3 +} = [Ar] 3 d^{1}, 4 s^{0}$
(one unpaired electron)
(3) ${TiO}_{2} \Rightarrow {Ti}^{4 +} = [Ar] 3 d^{0}, 4 s^{0}$
(no unpaired electron)
${CuSO}_{4} \Rightarrow {Cu}^{2 +} = [Ar] 3 d^{9}, 4 s^{0}$
(one unpaired electron)
(4) ${TiCl}_{3} \Rightarrow {Ti}^{3 +} = [Ar] 3 d^{1}, 4 s^{0}$
(one unpaired electron)
${NiCl}_{2} \Rightarrow {Ni}^{2 +} = [Ar] 3 d^{8}, 4 s^{0}$
(two unpaired electrons)
$∴ {CuCl}_{2}$ and ${TiCl}_{3}$ will have same magnetic moment.
$∴ μ = \sqrt{1 (1 + 2)} = \sqrt{3} = 1.73 BM$

AIIMS - 2009

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