Rank these transition metal ions in order of decreasing

Ответ:

foreverandalways80

03.07.2021

Chemistry

The given transition metal ions in order of decreasing number of unpaired electrons are as follows.

$Mn^{4+} V^{3+} = Ni^{2+} Fe^{3+} Cu^{+}$

Explanation:

In atomic orbitals, the distribution of electrons of an atom is called electronic configuration.

The electronic configuration in terms of noble gases for the given elements are as follows.

Atomic number of Fe is 26.

$Fe^{3+} - [Ar] 3d^{5}$

So, there is only 1 unpaired electron present in $Fe^{3+}$ .

Atomic number of Mn is 25.

$Mn^{4+} - [Ar]3d^{3}$

So, there are only 3 unpaired electrons present in $Mn^{4+}$ .

Atomic number of V is 23.

$V^{3+} - [Ar] 3d^{2}$

So, there are only 2 unpaired electrons present in $V^{3+}$ .

Atomic number of Ni is 28.

$Ni^{2+} - [Ar] 3d^{8}$

So, there will be 2 unpaired electrons present in $Ni^{2+}$ .

Atomic number of Cu is 29.

$Cu^{+} - [Ar] 3d^{10}$

So, there is no unpaired electron present in $Cu^{+}$ .

Therefore, given transition metal ions in order of decreasing number of unpaired electrons are as follows.

$Mn^{4+} V^{3+} = Ni^{2+} Fe^{3+} Cu^{+}$

Thus, we can conclude that given transition metal ions in order of decreasing number of unpaired electrons are as follows.

$Mn^{4+} V^{3+} = Ni^{2+} Fe^{3+} Cu^{+}$

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Ответ:

mystury

08.03.2022

Chemistry

Answer : The IUPAC name of Al_2S_3 is aluminum sulfide.

Explanation :

For formation of a neutral ionic compound, the charges on cation and anion must be balanced.

The cation is formed by loss of electrons by metals and anions are formed by gain of electrons by non metals.

Ionic compound : It is a type of compound that is made up of the ions of the two different elements in which one element is a metal and another element is a non-metal.

The nomenclature of ionic compounds is given by:

Positive is written first.The negative ion is written next and a suffix is added at the end of the negative ion. The suffix written is '-ide'.

The given compound is, Al_2S_3

Here aluminum is having an oxidation state of (+3) called as $Al^{3+}$ cation and sulfur $S^{2-}$ is an anion with oxidation state of (-2). Thus, they combine and their oxidation states are exchanged and written in simplest whole number ratios to give neutral Al_2S_3 called as aluminum sulfide.

Hence, the IUPAC name of Al_2S_3 is aluminum sulfide.

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Rank these transition metal ions in order of decreasing number of unpaired electrons. a. Fe^3
b. Mn^4+
c. V3+
d. Ni^2+
e. Cu^+

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Rank these transition metal ions in order of decreasing number of unpaired electrons. a. Fe^3 b. Mn^4+ c. V3+ d. Ni^2+ e. Cu^+

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Rank these transition metal ions in order of decreasing number of unpaired electrons. a. Fe^3
b. Mn^4+
c. V3+
d. Ni^2+
e. Cu^+