Answer:
Iron has 5 unpaired electrons in Fe⁺³ state.
Explanation:
Iron having atomic number 26 has following electronic configuration in neutral state.
Fe = 1s², 2s², 2p⁶, 3s², 3p⁶, 4s², 3d⁶
When Iron looses three electrons it attains +3 charge with following electronic configuration.
Fe⁺³ = 1s², 2s², 2p⁶, 3s², 3p⁶, 3d⁵
The five electrons in d-orbital exist in unpaired form as,
3(dz)¹, 3d(xz)¹, 3d(yz)¹, 3d(xy)¹, 3(dx²-y²)¹
Answer:
In metallic bonds, the valence electrons from the s and p orbitals of the interacting metal atoms delocalize. That is to say, instead of orbiting their respective metal atoms, they form a “sea” of electrons that surrounds the positively charged atomic nuclei of the interacting metal ions.
Explanation:
Answer: 12.18 u
Explanation: The average atomic mass of an element is calculated by taking the weighted average of the atomic masses of its stable isotopes.
In other words, each stable isotope will contribute to the average mass of the element proportionally to its abundance.
Answer:
48 g/mol
Explanation:
Step 1: Calculate the mass of the gas (m)
According to the law of conservation of mass, the mass of the solid before the decomposition must be equal to the sum of the masses of the solid residue and the gas
mSolid = mResidue + mGas
mGas = mSolid - mResidue = 4.73 g - 4.10 g = 0.63 g
Step 2: Convert 320 cm³ to L
We will use the conversion factor 1 L = 1000 cm³.
320 cm³ × 1 L/1000 cm³ = 0.320 L
Step 3: Calculate the moles of gas (n)
The gas is at room temperature (298.15 K) and room pressure (1 atm). We can calculate the moles of gas using the ideal gas equation.
P × V = n × R × T
n = P × V/R × T
n = 1 atm × 0.320 L/(0.0821 atm.L/mol.K) × 298.15 K = 0.0131 mol
Step 4: Calculate the molecular mass of the gas (M)
We will use the following expression.
M = m/n = 0.63 g/0.0131 mol = 48 g/mol
The given complex ion is as follow,
[Ru (CN) (CO)₄]⁻
Where;
[ ] = Coordination Sphere
Ru = Central Metal Atom = <span>Ruthenium
CN = Cyanide Ligand
CO = Carbonyl Ligand
The charge on Ru is calculated as follow,
Ru + (CN) + (CO)</span>₄ = -1
Where;
-1 = overall charge on sphere
0 = Charge on neutral CO
-1 = Charge on CN
So, Putting values,
Ru + (-1) + (0)₄ = -1
Ru - 1 + 0 = -1
Ru - 1 = -1
Ru = -1 + 1
Ru = 0
Result:
<span>Oxidation state of the metal species in each complex [Ru(CN)(CO)</span>₄]⁻ is zero.