The d- and f- Block Elements Test

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The d- and f- Block Elements Test - 15

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Weekly Quiz Competition

Question 1
1 / -0

A complex $${\left[ {M{F_6}} \right]^{ - x}}$$ has magnetic moment of 4.91 B.M. whereas $${\left[ {M{{(CN)}_6}} \right]^{ - x}}$$ has zero magnetic moment, metal ion can be :

A
$$C{o^{ + 2}}$$

B
$$M{n^{ + 2}}$$

C
$$F{e^{ + 2}}$$

D
$$F{e^{ + 3}}$$

Solution

$$\begin{array}{l} According\, to\, question............ \\ \, \, \, \, \, \, \, from\, data\, can\, \, be\, oncluded\, that\, \, first\, complez\, comtains\, 4\, unpaired\, electron\, as\, its\, magnetic\, moment\, is\, \\ \, 4.91\, BM\, \, magnetic\, moment\, (\, \mu )=\sqrt { n } (n+2),where\, \, n\, is\, no.of\, unpaired\, electron: \\ In\, high\, spin\, complex\, ,\, metal\, \, \, ion\, has\, 4\, unpaired\, electron\, and\, \, in\, low\, spin\, complex\, it\, \, has\, zero\, unpaired\, electron.... \\ so, \\ \, \, \, \, Ion\, \, is\, F{ e^{ +2 } },{ d^{ 6 } }t_{ 2g }^{ 4 }e_{ g }^{ 2 }\, \, is\, high\, spin,\, 4\, unpaired\, \, electron. \\ F{ e^{ +2 } },{ d^{ 6 } }t_{ 2g }^{ 4 }e_{ g }^{ 2 }\, \, low\, spin\, ,zero\, unpaired\, electron. \\ \, \, \, \, \, \, \, \, \, \therefore \, \, \, ion\, is\, \, \, F{ e^{ +2 } }. \\ so\, the\, \, correct\, ption\, \, is\, C. \end{array}$$
Question 2
1 / -0

The ions from among the following which are colourless are
(i) $$Ti^{4+}$$ (ii) $$Cu^{+}$$ (iii) $$Co^{3+}$$ (iv) $$Fe^{2+}$$

A
i and ii only

B
i, ii and iii

C
iii and iv

D
ii and iii

Solution

$$Ti^{4+} \Rightarrow [Ar] 3d^04s^0$$, No unpaired $$e^-$$
$$Cu^{+} \Rightarrow [Ar] 3d^{10}4s^0$$, No unpaired $$e^-$$
$$Co^{3+} \Rightarrow [Ar] 3d^64s^0$$, 4 unpaired $$e^-$$
$$Fe^{2+} \Rightarrow [Ar] 3d^64s^0$$, 4 unpaired $$e^-$$

$$Ti^{4+}$$ and $$Cu^{+}$$ are colourless.
Option A is correct.
Question 3
1 / -0

Actinoids comprises of elements from:

A
$$Th$$ and $$Lu$$

B
$$Th$$ and $$Lr$$

C
$$Ac$$ and $$Lu$$

D
$$Ac$$ and $$Lr$$

Solution

Actinium series starts with Actinium (Ac) and ends with Lawrencium(Lr).

Hence, the correct option is $$\text{D}$$
Question 4
1 / -0

The chemical similarity between boron and silicon mainly due to equal value of their:

A
electronegativity

B
nuclear charge

C
charge to (ionic radius)$$^{2}$$ ratio

D
atomic volume

Solution

Actually, it is because of diagonal relationship. Such relationship occurs because crossing and descending the periodic table have opposite effects. It is due to ion polarising power mainly. This is defined as charge divided by square of the radius.
Question 5
1 / -0

Beryllium resembles Aluminium in properties. This is mainly due to:

A
equal electro negativity values of elements

B
equal atomic volumes of the elements

C
equal electron affinity

D
equal nuclear charges in their atoms

Solution

Beryllium resembles Aluminium in properties. This is due to similar electronegativity, similar atomic size and they diagonal relationship.
Question 6
1 / -0

If the IP of hydrogen in its ground state is 2.18 x$$10^{-18}$$ J/atom, then the electron affinity of $$Li^{3+}$$ ion is :

A
$$-2.18\times$$$$10^{-18}$$J/atom

B
$$-6.54 \times$$$$10^{-18}$$J/atom

C
$$-3.488 \times$$$$10^{-18}$$J/atom

D
$$-1.962\times$$$$10^{-17}$$ J/atom

Solution

E.A of $$Li^{+3} = -$$I.P of $$Li^{+2}$$
I.P of $$Li^{+2} = $$I.P of $$H \dfrac{Z^2}{n^2}$$
Therefore, I.P of $$Li^{+2}$$=$$2.18 \times 10^{-18} J/atom \times 9 $$ as Z=3
Thus, E.A of $$Li^{+3} =-1.962 \times 10^{-17}$$ J/atom
Question 7
1 / -0

If the seventh period is completed, the atomic number of the last element would be:

A
$$118$$

B
$$112$$

C
$$107$$

D
$$120$$

Solution

As the seventh period will be completely filled its electronic configuration must be in the manner of noble gas configuration i.e., $$ns^2,\: np^6$$ because the last element in every period is a noble gas element.

Hence, the electronic configuration of the last element of the seventh period is (Rn) $$5f^{14} 6d^{10} 7s^27p^6$$ as the electronic configuration contains $$118$$ electrons.

The last element in the seventh period will have the atomic number $$118$$.
Question 8
1 / -0

The element with $$Z = 106$$ belongs to :

A
$$3^{rd}$$ period

B
$$ 5^{th}$$ period

C
$$7^{th}$$ period

D
$$6^{th}$$ period

Solution

$$Z=106$$ is seaborgium and its electronic configaration is $$(Rn) 5f^{14}6d^{4}7s^{2}$$. as the differenting electron enter into $$7s$$, 7 is the principle quantum number which represents the number of period to which the element belongs hence, the element belongs to 7th period.
Question 9
1 / -0

Most of the radio active elements are present in:

A
Lanthanoids

B
Actinoids

C
Representative elements

D
Second transition series

Solution

The actinoids are radioactive elements and the earlier members have relatively long half-lives.

The most common and known element is Uranium, which is used as nuclear fuel when its converted into plutonium, through a nuclear reaction.

Option B is correct.
Question 10
1 / -0

The element californium belongs to a family of :

A
actinide series

B
alkali metal family

C
alkaline earth family

D
lanthanide series
Solution
Hint: It belongs to f-block element.

Correct answer: $$A$$

Explanation for correct option :
In the periodic table, Actinoid element or the actinide element, any of a series of $$15$$ consecutive chemical elements from actinium to lawrencium (atomic numbers $$(89-103$$).
The atomic number of californium is $$98$$
Let's write an electronic configuration for this

its electronic configuration is $$Rn^{86} 7s^2 5f^{10}$$

Hence, it is an f-block element and as it is in the 7th period, it is a part of the actinide series.

Explanation for correct options :
Alkali and alkaline earth metals belong to s-block elements hence these both options are incorrect
Lanthanide series have elements $$57$$ to $$71$$ but californium has atomic number $$98$$, hence it not belongs to lanthanide series.

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The d- and f- Block Elements Test - 15

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The d- and f- Block Elements Test - 15

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Question 1

$$C{o^{ + 2}}$$

$$M{n^{ + 2}}$$

$$F{e^{ + 2}}$$

$$F{e^{ + 3}}$$

Solution

Question 2

i and ii only

i, ii and iii

iii and iv

ii and iii

Solution

Question 3

$$Th$$ and $$Lu$$

$$Th$$ and $$Lr$$

$$Ac$$ and $$Lu$$

$$Ac$$ and $$Lr$$

Solution

Question 4

electronegativity

nuclear charge

charge to (ionic radius)$$^{2}$$ ratio

atomic volume

Solution

Question 5

equal electro negativity values of elements

equal atomic volumes of the elements

equal electron affinity

equal nuclear charges in their atoms

Solution

Question 6

$$-2.18\times$$$$10^{-18}$$J/atom

$$-6.54 \times$$$$10^{-18}$$J/atom

$$-3.488 \times$$$$10^{-18}$$J/atom

$$-1.962\times$$$$10^{-17}$$ J/atom

Solution

Question 7

$$118$$

$$112$$

$$107$$

$$120$$

Solution

Question 8

$$3^{rd}$$ period

$$ 5^{th}$$ period

$$7^{th}$$ period

$$6^{th}$$ period

Solution

Question 9

Lanthanoids

Actinoids

Representative elements

Second transition series

Solution

Question 10

actinide series

alkali metal family

alkaline earth family

lanthanide series

Solution

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