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2 years ago

Which pair of elements form an ionic bond with

Chemistry

2 answers:

Vesna [10]2 years ago

5 0

Answer:

It's only A

Explanation:

Potassium (K) is a metal and Chlorine (Cl) is a nonmetal, and ionic compounds are formed between metals and nonmetals. So KCl forms ionic bonds.

Hydrogen (H) and Chlorine (Cl) are both nonmetals, and only covalent bonds form between nonmetals. (Only exception is when its an acid, then it's ionic)

Also, the question indicated which "pair" and not "pairs" meaning that there is only one answer, and I'm more likely to go with A rather than D.

KATRIN_1 [288]2 years ago

3 0

Answer:

Yes Both A) KCL and D) HCL are correct...

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The NaCl crystal structure consists of alternating Na⁺ and Cl⁻ ions lying next to each other in three dimensions. If the Na⁺ rad

Yuliya22 [10]

Answer : The radii of the two ions Cl⁻ ion and Na⁺ ion is, 181 and 102 pm respectively.

Explanation :

As we are given that the Na⁺ radius is 56.4% of the Cl⁻ radius.

Let us assume that the radius of Cl⁻ be, (x) pm

So, the radius of Na⁺ = $x\times \frac{56.4}{100}=(0.564x)pm$

In the crystal structure of NaCl, 2 Cl⁻ ions present at the corner and 1 Na⁺ ion present at the edge of lattice.

Thus, the edge length is equal to the sum of 2 radius of Cl⁻ ion and 2 radius of Na⁺ ion.

Given:

Distance between Na⁺ nuclei = 566 pm

Thus, the relation will be:

$2\times \text{Radius of }Cl^-+2\times \text{Radius of }Na^+=\text{Distance between }Na^+\text{ nuclei}$

$2\times x+2\times 0.564x=566$

$2x+1.128x=566$

$3.128x=566$

$x=180.9\approx 181pm$

The radius of Cl⁻ ion = (x) pm = 181 pm

The radius of Na⁺ ion = (0.564x) pm = (0.564 × 181) pm =102.084 pm ≈ 102 pm

Thus, the radii of the two ions Cl⁻ ion and Na⁺ ion is, 181 and 102 pm respectively.

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