Tag: the d-and f-block elements

Questions Related to the d-and f-block elements

Which of the following is true regarding derivation of the name of californium?

chemistry d- and f-block elements actinoids the actinoids the d-and f-block elements comparison of lanthanoids and actinoids

  1. The name of californium was derived from the name of the place.

  2. The name of californium was derived from its color.

  3. The name of californium was derived from the name of the scientist who discovered it.

  4. The name of californium was derived from mythological character name.

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Correct Option: A
Explanation:

The name of californium was derived from the name of the state and University of California.

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

chemistry d- and f-block elements actinoids the actinoids the d-and f-block elements comparison of lanthanoids and actinoids

  1. equal electro negativity values of elements

  2. equal atomic volumes of the elements

  3. equal electron affinity

  4. equal nuclear charges in their atoms

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Correct Option: A
Explanation:

Beryllium resembles Aluminium in properties. This is due to similar electronegativity, similar atomic size and they diagonal relationship.

Amphoteric behaviour is shown by the oxides of :

chemistry d- and f-block elements actinoids the actinoids the d-and f-block elements comparison of lanthanoids and actinoids

  1. $Al$ and $Ca$

  2. $Pb$ and $Ba$

  3. $Cr$ and $Mg$

  4. $Sn$ and $Zn$

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Correct Option: D
Explanation:

Amphoteric oxides can behave as an acid as well as base. They react with both acids and bases to form salts. Tin oxide although doesn't dissolve in water but it is amphoteric in nature. Zinc oxide is also amphoteric in nature.

$ZnO+HCl \rightarrow ZnCl _2+H _2O$
$ZnO+NaOH \rightarrow Na _2ZnO _2+H _2O$
$SnO _2+HCl \rightarrow SnCl _4+H _2O$
$SnO _2+NaOH+H _2O \rightarrow Na _2[Sn(OH) _6]$

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 :

chemistry d- and f-block elements actinoids the actinoids the d-and f-block elements comparison of lanthanoids and actinoids

  1. $-2.18\times$$10^{-18}$J/atom

  2. $-6.54 \times$$10^{-18}$J/atom

  3. $-3.488 \times$$10^{-18}$J/atom

  4. $-1.962\times$$10^{-17}$ J/atom

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Correct Option: D
Explanation:

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

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

chemistry d- and f-block elements actinoids the actinoids the d-and f-block elements comparison of lanthanoids and actinoids

  1. $118$

  2. $112$

  3. $107$

  4. $120$

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Correct Option: A
Explanation:

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 noble gas element. Hence, the electronic configuration of last element of 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 $118$.

Stable electronic configuration of a transition metal is:

chemistry d- and f-block elements actinoids the actinoids the d-and f-block elements comparison of lanthanoids and actinoids

  1. $1s^{2} 2s^{2}2p^{6} 3s^{2} 3p^{6} 3d^{9} 4s^{2}$

  2. $1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 4s^{2} 3d^{5}$

  3. $1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 4s^{2}$

  4. $1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 4s^{2} 3d^{10} 4p^{6}$

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Correct Option: B
Explanation:

Generally the elements having  half filled and completely filled outer electronic configurations are stable  elements,here in given options the configuration $1s^22s^22p^63s^23p^64s^23d^5$ contains half filled outer electronic configuration $(3d^5)$.


Hence option B is correct.

The element with $Z = 106$ belongs to :

chemistry d- and f-block elements actinoids the actinoids the d-and f-block elements comparison of lanthanoids and actinoids

  1. $3^{rd}$ period

  2. $ 5^{th}$ period

  3. $7^{th}$ period

  4. $6^{th}$ period

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Correct Option: C
Explanation:

$Z=106$ 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.

An atom of element has $2K, 8L$ and $3M$ electrons. Then the element belongs to :

chemistry d- and f-block elements actinoids the actinoids the d-and f-block elements comparison of lanthanoids and actinoids

  1. IA group

  2. IIA group

  3. IIIA group

  4. IVA group

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Correct Option: C
Explanation:

The total number of electrons present in all the shells are $2+8+3=13$, hence the atomic number of element is $13$ and the electronic configuration would be $1s^22s^22p^63s^23p^1 $. Element belongs to $|||A$ group.

Most of the man-made synthetic elements occur:

chemistry d- and f-block elements actinoids the actinoids the d-and f-block elements comparison of lanthanoids and actinoids

  1. in actinide series

  2. in lanthanide series

  3. in coinage metals

  4. in alkaline earth metals

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Correct Option: A

Maximum oxidation state of uranium and plutonium are respectively:

chemistry d- and f-block elements actinoids the actinoids the d-and f-block elements comparison of lanthanoids and actinoids

  1. $4$ and $6$

  2. $5$ and $6$

  3. $6$ and $7$

  4. $6$ and $8$

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Correct Option: C
Explanation:
Solution:- (C) $6$ and $7$
Uranium has maximum oxidation number $= +6$ 
Plutonium has maximum oxidation number $= +7$