Answer
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Hint: We know that the distance between the valence shell and the nucleus's centre is termed as atomic radius. The atomic radius is expressed in three different forms namely, covalent radii, van der Waals radii and metallic radii depending on the type of bonding between the atoms.
Complete answer:
Let's discuss the variation of atomic radius in the periodic table. Along a period, the atomic radii of elements decreases on moving from left to right and down the group, the atomic radii of elements increases.
Let's discuss the reason for such variation along the periods. On moving from left to right in a period, there is an increase of nuclear charge by one unit and one electron is also added to the valence shell at the same time. The increase of nuclear charge from left to right results in more attractions of electrons towards the nucleus. Therefore, atomic size decreases.
Now, we discuss the variation of atomic size from down the group. On moving down the group, there is an increase of principal quantum number (shells). Thus, the sizes of atoms are expected to increase. Moreover, the atomic number or nuclear charge also increases which must decrease the atomic size. But, the effect of nuclear charge is more pronounced than the effect of increase in nuclear charge. Therefore, atomic radius increases down the group.
Here, we have to identify the largest atomic radii among Se, Br, Te and I. Selenium (Se) and Tellurium (I) belong to the same group 16 and 4th and 5th period respectively. So, Te is larger Se.
Bromine (Br) and Iodine (I) belong to group 17 and 4th and 5th period respectively. So, I is larger than Br.
Now, we have to compare the size between Te and I. Te and I both belong to the 5th period and group 16 and 17 respectively. So, the size or atomic radius Te is largest.
So, the correct answer is Option C.
Note: In homoatomic molecules, the covalent radius is defined as the half of internuclear distance of the two similar atoms bonded by a single covalent bond. In case of double covalent or triple covalent bond, the half of internuclear distance does not represent covalent radius.
Complete answer:
Let's discuss the variation of atomic radius in the periodic table. Along a period, the atomic radii of elements decreases on moving from left to right and down the group, the atomic radii of elements increases.
Let's discuss the reason for such variation along the periods. On moving from left to right in a period, there is an increase of nuclear charge by one unit and one electron is also added to the valence shell at the same time. The increase of nuclear charge from left to right results in more attractions of electrons towards the nucleus. Therefore, atomic size decreases.
Now, we discuss the variation of atomic size from down the group. On moving down the group, there is an increase of principal quantum number (shells). Thus, the sizes of atoms are expected to increase. Moreover, the atomic number or nuclear charge also increases which must decrease the atomic size. But, the effect of nuclear charge is more pronounced than the effect of increase in nuclear charge. Therefore, atomic radius increases down the group.
Here, we have to identify the largest atomic radii among Se, Br, Te and I. Selenium (Se) and Tellurium (I) belong to the same group 16 and 4th and 5th period respectively. So, Te is larger Se.
Bromine (Br) and Iodine (I) belong to group 17 and 4th and 5th period respectively. So, I is larger than Br.
Now, we have to compare the size between Te and I. Te and I both belong to the 5th period and group 16 and 17 respectively. So, the size or atomic radius Te is largest.
So, the correct answer is Option C.
Note: In homoatomic molecules, the covalent radius is defined as the half of internuclear distance of the two similar atoms bonded by a single covalent bond. In case of double covalent or triple covalent bond, the half of internuclear distance does not represent covalent radius.
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