
La (lanthanum) having atomic number 57 is a member of:
(a)- s-block element
(b)- p-block element
(c)- d-block element
(d)- f-block element
Answer
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Hint: The block of the element is decided on the basis of the electronic configuration of the element, the electron which enters the last sub-shell decides the block of the element. Lanthanum is having an atomic number 57 so it has 57 electrons and these 57 electrons will be filled according to the increasing energies of the orbitals.
Complete step by step answer:
- The block of the element is decided on the basis of the electronic configuration of the element, the electron which enters the last sub-shell decides the block of the element.
- The periodic table is the arrangement of elements according to their increasing number of electrons and these electrons are filled in the orbits according to their increasing energies. There are various shells in the atoms which are denoted by K, L, M, N, etc, which represents the specific energy of the shell. The shell K represents 1, the shell L represents 2, the shell M represents 3, the shell N represents 4. Each shell has specific sub-shells which are s, p, d, f, etc.
- Lanthanum is having an atomic number 57 so it has 57 electrons and these 57 electrons will be filled according to the increasing energies of the orbitals.
- There are many orbitals that are placed according to their increasing order of energies. The order is $\text{1s 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p 6s 4f 5d 6p 7s 5f 6d 7p}$.
- So, the electronic configuration will be:
$\text{1}{{\text{s}}^{2}}\text{ 2}{{\text{s}}^{2}}\text{ 2}{{\text{p}}^{6}}\text{ 3}{{\text{s}}^{2}}\text{ 3}{{\text{p}}^{6}}\text{ 4}{{\text{s}}^{2}}\text{ 3}{{\text{d}}^{10}}\text{ 4}{{\text{p}}^{6}}\text{ 5}{{\text{s}}^{2}}\text{ 4}{{\text{d}}^{10}}\text{ 5}{{\text{p}}^{6}}\text{ 6}{{\text{s}}^{2}}\text{ 5}{{\text{d}}^{1}}$
Therefore, the last electron is in d-subshell, hence the block will be d-block.
The correct option is option “C” .
Note: Some properties of lanthanum are soft, ductile, and it is a silvery-white metal. It is a very soft metal and it can be cut even with a knife and when it is exposed to air, it starts to tarnish.
Complete step by step answer:
- The block of the element is decided on the basis of the electronic configuration of the element, the electron which enters the last sub-shell decides the block of the element.
- The periodic table is the arrangement of elements according to their increasing number of electrons and these electrons are filled in the orbits according to their increasing energies. There are various shells in the atoms which are denoted by K, L, M, N, etc, which represents the specific energy of the shell. The shell K represents 1, the shell L represents 2, the shell M represents 3, the shell N represents 4. Each shell has specific sub-shells which are s, p, d, f, etc.
- Lanthanum is having an atomic number 57 so it has 57 electrons and these 57 electrons will be filled according to the increasing energies of the orbitals.
- There are many orbitals that are placed according to their increasing order of energies. The order is $\text{1s 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p 6s 4f 5d 6p 7s 5f 6d 7p}$.
- So, the electronic configuration will be:
$\text{1}{{\text{s}}^{2}}\text{ 2}{{\text{s}}^{2}}\text{ 2}{{\text{p}}^{6}}\text{ 3}{{\text{s}}^{2}}\text{ 3}{{\text{p}}^{6}}\text{ 4}{{\text{s}}^{2}}\text{ 3}{{\text{d}}^{10}}\text{ 4}{{\text{p}}^{6}}\text{ 5}{{\text{s}}^{2}}\text{ 4}{{\text{d}}^{10}}\text{ 5}{{\text{p}}^{6}}\text{ 6}{{\text{s}}^{2}}\text{ 5}{{\text{d}}^{1}}$
Therefore, the last electron is in d-subshell, hence the block will be d-block.
The correct option is option “C” .
Note: Some properties of lanthanum are soft, ductile, and it is a silvery-white metal. It is a very soft metal and it can be cut even with a knife and when it is exposed to air, it starts to tarnish.
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