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Marianna [84]
3 years ago
7

The d-metals iron, copper, and manganese form cations with different oxidation states. For this reason, they are found in many o

xidoreductases and in several proteins of oxidative phosphorylation and photosynthesis. Explain why many d-metals form cations with different oxidation states.
Chemistry
2 answers:
kicyunya [14]3 years ago
8 0

Answer:

==> Presence of unpaired electrons.

==> Electrons in ns and electrons in (n-1)d orbitals are used to form bonds because of their similarities in energies.

Explanation:

The d- metals are also known as the transition metals with metal elements such as the ones given in the question, that is; Copper, iron and Manganese. Others include chromium, Nickel, Cobalt and so on.

The reason behind many d-metals forming cations with different oxidation states is mainly because of the unpaired electrons. Because Unpaired electrons are unstable they tends to attract or bond with other species chemically.

Also, many d-metals form cations with different oxidation states because Electrons in ns and electrons in (n-1)d orbitals are used to form bonds because of their similarities in energies.

choli [55]3 years ago
4 0

Answer:

They have electrons in their 3d- and 4s-orbital for bond formation.

Explanation:

d- metals or transition metal are metal which form ion with partially filled d-orbital. Examples are iron and manganese.

The metals have 2 electrons in their 4s orbital. If only this is used for bonding, they will form compounds where they have oxidation State of +2 as seen in MnO.

If two 4s and one of 3d electrons are used, oxidation state of +3 is formed as seen in FeCl3.

If two 2s electron I used with two 3d electrons, compound with oxidation state of +4 is formed as seen in MnO2

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<h3>Answer:</h3>

1000 g CCl₄

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right

<u>Chemistry</u>

<u>Atomic Structure</u>

  • Reading a Periodic Table
  • Using Dimensional Analysis
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<h3>Explanation:</h3>

<u>Step 1: Define</u>

3.93 × 10²⁴ molecules CCl₄

<u>Step 2: Identify Conversions</u>

Avogadro's Number

Molar Mass of C - 12.01 g/mol

Molar Mass of Cl - 35.45 g/mol

Molar Mass of CCl₄ - 12.01 + 4(35.45) = 153.81 g/mol

<u>Step 3: Convert</u>

  1. Set up:                              \displaystyle 3.93 \cdot 10^{24} \ molecules \ CCl_4(\frac{1 \ mol CCl_4}{6.022 \cdot 10^{23} \ molecules \ CCl_4})(\frac{153.81 \ g \ CCl_4}{1 \ mol \ CCl_4})
  2. Multiply:                                                                                                             \displaystyle 1003.77 \ g \ CCl_4

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 3 sig figs.</em>

1003.77 g CCl₄ ≈ 1000 g CCl₄

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<h3><u>For the given chemical equation:</u></h3>

Cu + 2AgNO₃ →  2Ag + Cu(NO₃)₂

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<h3><u>Oxidising and Reducing agents</u></h3>
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