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Katen [24]
3 years ago
7

An aluminum cation is smaller than the aluminum atom because it has fewer occupied energy levels. True false

Chemistry
2 answers:
Mars2501 [29]3 years ago
7 0

Answer: The given statement is true.

Explanation:

Atomic number of aluminium is 13 and in order to attain stability aluminium loses 3 electrons. As a result, neutral aluminium atom converts into Al^{3+} ion.

Thus, due to loss of electrons aluminium cation is smaller than neutral aluminium atom.

Therefore, the statement an aluminum cation is smaller than the aluminum atom because it has fewer occupied energy levels, is true.

butalik [34]3 years ago
3 0

The molecules or atoms that are formed by gain or loss of one or more valence electrons are said to be ions.

When atom loss one or more valence electrons, results in formation of cation whereas when atom gain one or more valence electrons, then formation of anion occurs. Cations carry positive charge and anions carry negative charge.

In general, cations are smaller than the neutral atoms from which they are formed and anions are larger than the neutral atoms.

As cations are smaller than the related neutral atoms because the valence electrons are lost which are farthest away from the nucleus. After that, taking more electrons distant from the cation results in reduction of radius of the ion.

Thus, aluminium cation consist of few electrons which results in fewer occupied energy levels by the electrons further results in reduction of radius i.e. smaller size.

Hence, given statement is true i.e. aluminium atom is larger than the aluminium cation as cation has fewer occupied energy levels.



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Explanation:

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3 years ago
A person applies a force of 200 N over 1.5 m to a jack. The jack exerts a 1000-N force on a car a distance of 0.02 m. What is th
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The efficiency rating of the jack is 0.067.

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3 years ago
Indicate whether the following balanced equations involve oxidation-reduction. Check all that apply. Check all that apply. 2H2SO
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Answer :  The balanced equations involve oxidation-reduction are:

(a) 2H_2SO_4(aq)+2NaBr(s)\rightarrow Br_2(l)+SO_2(g)+Na_2SO_4(aq)+2H_2O(l)

(b) 3SO_2(g)+2HNO_3(aq)+2H_2O(l)\rightarrow 3H_2SO_4(aq)+2NO(g)

(c) NaI(aq)+3HOCl(aq)\rightarrow NaIO_3(aq)+3HCl(aq)

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Redox reaction or Oxidation-reduction reaction : It is defined as the reaction in which the oxidation and reduction reaction takes place simultaneously.

Oxidation reaction : It is defined as the reaction in which a substance looses its electrons. In this, oxidation state of an element increases. Or we can say that in oxidation, the loss of electrons takes place.

Reduction reaction : It is defined as the reaction in which a substance gains electrons. In this, oxidation state of an element decreases. Or we can say that in reduction, the gain of electrons takes place.

(a) The given chemical reaction is:

2H_2SO_4(aq)+2NaBr(s)\rightarrow Br_2(l)+SO_2(g)+Na_2SO_4(aq)+2H_2O(l)

This reaction involve oxidation-reduction reaction because the oxidation state bromine changes from (-1) to (0) which shows oxidation and sulfur changes from (+6) to (+4) which shows reduction.

(b) The given chemical reaction is:

3SO_2(g)+2HNO_3(aq)+2H_2O(l)\rightarrow 3H_2SO_4(aq)+2NO(g)

This reaction involve oxidation-reduction reaction because the oxidation state sulfur changes from (+4) to (+6) which shows oxidation and nitrogen changes from (+5) to (+2) which shows reduction.

(c) The given chemical reaction is:

NaI(aq)+3HOCl(aq)\rightarrow NaIO_3(aq)+3HCl(aq)

This reaction involve oxidation-reduction reaction because the oxidation state iodine changes from (-1) to (+5) which shows oxidation and chlorine changes from (+5) to (-1) which shows reduction.

(d) The given chemical reaction is:

PBr_3(l)+3H_2O(l)\rightarrow H_3PO_3(aq)+3HBr(aq)

This reaction does not involve oxidation-reduction reaction because the oxidation state of element present on reactant and product side are same.

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