
Non-metals that has a metallic lustre and sublimes on heating is:
(a) chlorine
(b) bromine
(c) oxygen
(d) iodine
Answer
567.6k+ views
Hint: Metallic lustre is basically a property shown by metals where the surface of the elements can reflect light, and it can be polished also to give a shiny surface. Sublime property is the transformation of solid state directly into gaseous state.
Complete step by step answer:
(1) As the question is saying that we have to find a non-metal with the given properties, so, we should check which of the given four options are non-metal, so that we can eliminate that option which is a metal.
(2) For checking whether an element is a metal or nonmetal, we have to find the number of electrons in the outermost shell which is also called the valence shell of the atom. This can be done by writing the electronic configuration of the elements. Elements with a number of electrons between 5-8 are non-metals.
(3) For option (a) chlorine, its atomic number is 17, so, its electronic configuration will be ${\left[ {{\text{Ne}}} \right]_{10}}3{s^2}3{p^5}$. Its outermost shell is ${\text{3p and 3s}}$, so it has 7 electrons in the outermost shell. Hence is a non-metal.
For option (b) bromine, its atomic number is 35, so, its electronic configuration will be ${\left[ {{\text{Ar}}} \right]_{18}}4{s^2}3{d^{10}}4{p^5}$. Its outermost shell is ${\text{4p and 4s}}$, so it has 7 electrons in the outermost shell. Hence is a non-metal.
For option (c) oxygen, its atomic number is 8, so its electronic configuration will be $1{s^2}2{s^2}2{p^4}$. Its outermost shell is ${\text{2p and 2s}}$, so it has 6 electrons in the outermost shell. Hence is a non-metal.
For option (d) Iodine, its atomic number is 53, so, its electronic configuration will be ${\left[ {{\text{Kr}}} \right]_{18}}5{s^2}4{d^{10}}5{p^5}$. Its outermost shell is ${\text{5p and 5s}}$, so it has 7 electrons in the outermost shell. Hence is a non-metal.
(4) Among all the four non-metals Iodine is the one which can show the property of metallic lustre because electrons in Iodine are loosely bound, so they can oscillate to show the lustre. Moreover, the partial pressure of Iodine is very low due to which it undergoes a spontaneous reaction of sublimation where it is changed from its solid phase to gaseous phase without going into the liquid phase.
Hence, option (d) Iodine is the correct answer.
Note:
After writing the electronic configuration of elements like Iodine, carefully choose the outermost shell carefully because in Iodine, last electron goes into the ${\text{4d}}$ but the outermost shell is ${\text{5p and 5s}}$.
Complete step by step answer:
(1) As the question is saying that we have to find a non-metal with the given properties, so, we should check which of the given four options are non-metal, so that we can eliminate that option which is a metal.
(2) For checking whether an element is a metal or nonmetal, we have to find the number of electrons in the outermost shell which is also called the valence shell of the atom. This can be done by writing the electronic configuration of the elements. Elements with a number of electrons between 5-8 are non-metals.
(3) For option (a) chlorine, its atomic number is 17, so, its electronic configuration will be ${\left[ {{\text{Ne}}} \right]_{10}}3{s^2}3{p^5}$. Its outermost shell is ${\text{3p and 3s}}$, so it has 7 electrons in the outermost shell. Hence is a non-metal.
For option (b) bromine, its atomic number is 35, so, its electronic configuration will be ${\left[ {{\text{Ar}}} \right]_{18}}4{s^2}3{d^{10}}4{p^5}$. Its outermost shell is ${\text{4p and 4s}}$, so it has 7 electrons in the outermost shell. Hence is a non-metal.
For option (c) oxygen, its atomic number is 8, so its electronic configuration will be $1{s^2}2{s^2}2{p^4}$. Its outermost shell is ${\text{2p and 2s}}$, so it has 6 electrons in the outermost shell. Hence is a non-metal.
For option (d) Iodine, its atomic number is 53, so, its electronic configuration will be ${\left[ {{\text{Kr}}} \right]_{18}}5{s^2}4{d^{10}}5{p^5}$. Its outermost shell is ${\text{5p and 5s}}$, so it has 7 electrons in the outermost shell. Hence is a non-metal.
(4) Among all the four non-metals Iodine is the one which can show the property of metallic lustre because electrons in Iodine are loosely bound, so they can oscillate to show the lustre. Moreover, the partial pressure of Iodine is very low due to which it undergoes a spontaneous reaction of sublimation where it is changed from its solid phase to gaseous phase without going into the liquid phase.
Hence, option (d) Iodine is the correct answer.
Note:
After writing the electronic configuration of elements like Iodine, carefully choose the outermost shell carefully because in Iodine, last electron goes into the ${\text{4d}}$ but the outermost shell is ${\text{5p and 5s}}$.
Recently Updated Pages
Why are manures considered better than fertilizers class 11 biology CBSE

Find the coordinates of the midpoint of the line segment class 11 maths CBSE

Distinguish between static friction limiting friction class 11 physics CBSE

The Chairman of the constituent Assembly was A Jawaharlal class 11 social science CBSE

The first National Commission on Labour NCL submitted class 11 social science CBSE

Number of all subshell of n + l 7 is A 4 B 5 C 6 D class 11 chemistry CBSE

Trending doubts
Differentiate between an exothermic and an endothermic class 11 chemistry CBSE

10 examples of friction in our daily life

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

State the laws of reflection of light

Explain zero factorial class 11 maths CBSE

