What Is The Oxidation State Of Each Element In Febr2? ? Ferrous Bromide (Febr2) Oxidation Number

The oxidation states of the elements Fe and Br needs to be determined.Fe is a transition element therefore is capable of multiple oxidation states. an ionic compound is made with Fe as the cation and halide ion as anionBr⁻ which is a halogen has a charge of -1. In ions the charge of the ion is equal to the oxidation state of the ion. Br⁻ has oxidation state of -1. FeBr₂ is a compound therefore net charge is 0.the sum of the products of the oxidation states of the individual elements by the number of atoms should add upto the net chargeoxidation number of Fe cation + (oxidation number of Br⁻ x 2) = 0say the oxidation state of Fe is X and oxidation number of Br⁻ is -1X + (2 x -1)= 0X – 2 = 0X = 2this means that oxidation number of Fe is +2and oxidation number of Br is -1

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In standardization of NaOH solution, a student found that 25.55cm3 of base neutralized exactly 21.35cm3 of 0.12M HCl. Find the m



We”ll begin by writing a balanced equation for the reaction between HCl and NaOH. This is illustrated below:

HCl + NaOH —> NaCl + H2O

From the above equation, we obtained:

nA (mole of the acid) = 1

nB (mole of the base) = 1

Data obtained from the question include:

Ma (Molarity of the acid ) = 0.12M

Va (volume of the acid) = 21.35cm3

Mb (Molarity of the base) =?

Vb (volume of the base) = 25.55cm3

Using the equation MaVa/MbVb = nA/nB, the molarity of the base can be obtained as follow:

MaVa/MbVb = nA/nB

0.12 x 21.35 / Mb x 25.55 = 1

Cross multiply to express in linear form.

Mb x 25.55 = 0.12 x 21.35

Divide both side by 25.55

Mb = (0.12 x 21.35)/25.55

Mb = 0.1M

Therefore, the molarity of the base (NaOH) is 0.1M

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Group 1A contains the most electropositive metal whereas Group VIIA contains the most electronegative non-metal. The valence shell of group 1A is made up of a single electron but those of VIIA contains seven electrons. Based on the valence electrons, Group 1A will readily lose an electron in order to complete its octet with an oxidation state of +1Group VIIA are highly reactive group of elements with a dire need to gain an electron to complete their octet configuration. The most reactive metal is Francium whereas Fluorine is the most reactive non-metal.

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Alkali metal and halogens


C4H10 + 6.5 O2 —-> 4CO2 + 5H2O 2C4H10 + 13 O2 —-> 8CO2 + 10H2O 1. Count the C on the left (4), put a 4 where the C on the right. 2. Count the H on the left (1), you have two on the right, so you multimply this two by 5. Put the 5 in front of the H2O 3. Count the O on the right. You have 4*2 + 5 = 13. You have two on the left, so you need 6.5 on the left. 4. Now multiply everything on the equation by two so you have nice integer numbers. 5. check you have the same amount of everything on each side. Example C: left 8, right 8, etc.I hope this helps. 🙂

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CuSO₄.H₂O = 1

CuSO₄.3H₂O = 3

CuSO₄.5H₂O = 5

CuSO₄7.H₂O = 7

CuSO₄.9H₂O = 9


Some salts as CuSO₄ are presented in the hydratated form to give some stability in their uses.

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