If the ionic radius of M+
and X–
are respectively 135 pm and 209 pm, the expected value of atomic
radius of metal (M) and non metal (X) are respectively (X and M are elements of same period)
A) 180 pm & 90 pm
B) 135 pm & 209 pm
C) 90 pm & 180 pm
D) Same radius which is average of 135 pm & 209 pm
PLZ EXPLAIN
Answers
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If the ionic radii of K^+ and F^- are about 1.34 A each, then the expected values of atomic radii of K and F should be respectively:
Answered by
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Answer:
The expected value of the atomic radius of = .
The expected value of the atomic radius of = .
Therefore, option A) and is correct.
Explanation:
Data given,
The ionic radius of =
The ionic radius of =
The expected value of the atomic radius of =?
The expected value of the atomic radius of =?
As we know,
- The atomic radius of the positive ion ( cation ) of an element is larger than the ionic radius. ( because cations are always shorter than the parent atoms ).
- Also, the more the positive ions pull the electrons closer. So the cation or the positive ion is smaller than the atom.
Similarly,
- The atomic radius of the negative ion ( anion ) of an element is smaller than the ionic radius. ( because anions are always larger than the parent atoms ).
- Also, the more the negative charge on the ion, the more the electrons repel each other.
Thus, the atomic radius of should be greater than .
And the atomic radius of should be less than .
After applying these conditions, let us check the options:
A) and
- = greater than .
- = less than .
B) and
- = equals to .
- = equals to .
C) and
- = less than .
- = less than .
D) The same radius which is an average of & .
No, the radius of the ions cannot be the same.
Thus, from the above options, only option A) matches the criterion,
Therefore,
- The expected value of the atomic radius of = .
- The expected value of the atomic radius of = .
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