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AlekseyPX
3 years ago
7

Which atom attracts electrons more strongly?

Chemistry
2 answers:
lesya [120]3 years ago
3 0
Bromine attracts electrons more strongly. Cesium is In fact the least electro negative element.

Sodium is more likely to lose an electron because is is less electro negative. Strong electronegativity make the element want more electrons. Sodium has loose electrons with a lower electronegativity so it gives it up easier.
makvit [3.9K]3 years ago
3 0

Explanation:

1.    Cesium (Cs) is a group 1 element and it has electronic configuration [Xe] 6s^{1}.

On the other hand, bromine is a group 7 element and it has electronic configuration [Ar] 3d^{10}4s^{2}40^{5}.

Since, Cs is a metal so it will readily lose 1 electron but bromine being a non-metal needs 1 electron to completely fill its orbital.

Hence, bromine atom will attract electrons more strongly.

2.   Sodium (Na) is a a group 1 element and it has electronic configuration 1s^{2}2s^{2}2p^{6}3s^{1}.

Whereas, chlorine is a group 7 element and it has electronic configuration 1s^{2}2s^{2}2p^{6}3s^{2}3p^{5}.

So, sodium is a metal hence, it readily loses its one electron and chlorine being a non-metal will attract one electron in order to completely fill its orbital.

Thus, we can conclude that sodium (Na) atom loses an electron more readily.

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6 0
3 years ago
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7 0
2 years ago
When electrons move closer to a more electronegative atom, what happens? the more electronegative atom is _____?
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3 years ago
Which TWO statements describe the ocean floor giving brainliest pls help explain ur answer
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8 0
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The reaction is proceeding at a rate of 0.0080 Ms-1 in 50.0 mL of solution in a system with unknown concentrations of A and B. W
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Explanation:

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rate = k[A][B]²

If you are comparing two rates,

\dfrac{\text{rate}_{2}}{\text{rate}_{1}} = \dfrac{k_{2}\text{[A]}_2[\text{B]}_{2}^{2}}{k_{1}\text{[A]}_1[\text{B]}_{1}^{2}}= \left (\dfrac{\text{[A]}_{2}}{\text{[A]}_{1}}\right ) \left (\dfrac{\text{[B]}_{2}}{\text{[B]}_{1}}\right )^{2}

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\dfrac{\text{[A]}_{2}}{\text{[A]}_{1}} = \dfrac{1}{2}\text{ and }\dfrac{\text{[B]}_{2}}{\text{[B]}_{1}}= \dfrac{1}{2}

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8 0
3 years ago
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