What does a metal's position in an activity series indicate about the chemical properties of that particular metal?

1 answer:

3 0

The closer to the top the metal is in the list, the more active the metal is and the stronger a reducing agent the metal is. When two different metals are involved in a redox reaction, the metal higher in the list will be oxidized and give up electrons that will reduce the cation of the less active metal.

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3. What noble gas would be part of the electron configuration notation for Mn?

shusha [124]

Answer:

Argon {Ar}

Explanation:

The noble gas used for a condensed electron configuration is the one before the element which you are configuring.

In this case, the element (Mn) is manganese

The noble gas that is before this element is Argon which is the row above it

so your configuration would be {Ar} 3d^5 4s^2

7 0

2 years ago

Write the empirical formula for at least four ionic compounds that could be formed from the following ions:

CaHeK987 [17]

Answer:

Fe(CN)₂, FeCO₃, Pb(CN)₄, Pb(CO₃)₂

Explanation:

Cations (positively charged ions) can only form ionic bonds with anions (negatively charged ions). However, you can't just simply put one cation and one anion together to form a compound. Each compound needs to been neutral, or have an overall charge of 0. When cations and anions do not have charges that perfectly cancel, you need to modify the amount of each ion in the compound.

1.) Fe(CN)₂
-----> Fe²⁺ and CN⁻

-----> +2 + (-1) + (-1) = 0

2.) FeCO₃

-----> Fe²⁺ and CO₃²⁻

-----> +2 + (-2) = 0

3.) Pb(CN)₄

-----> Pb⁴⁺ and CN⁻

-----> +4 + (-1) + (-1) + (-1) + (-1) = 0

4.) Pb(CO₃)₂

-----> Pb⁴⁺ and CO₃²⁻

-----> +4 +(-2) + (-2) = 0

5 0

1 year ago

What orbitals are used to form the 10 sigma bonds in propane (ch3ch2ch3)? Label each atom with the appropriate hybridization. Dr

SVETLANKA909090 [29]

Mixing of pure orbitals having nearly equal energy to form equal number of completely new orbitals is said to be hybridization.

For the compound, $CH_3CH_2CH_3$ the electronic configuration of the atoms, carbon and hydrogen are:

Carbon (atomic number=6): In ground state= $1s^{2}2s^{2}2p^{2}$

In excited state: $1s^{2}2s^{1}2p^{3}$

Hydrogen (atomic number=1): $1s^{1}$

All the bonds in the compound is single bond( $\sigma$ -bond) that is they are formed by head on collision of the orbitals.

The structure of the compound is shown in the image.

The Carbon-Hydrogen bond is formed by overlapping of s-orbital of hydrogen to p-orbital of carbon.

In order to complete the octet the required number of electrons for carbon is 4 and for hydrogen is 1. So, the electron in $1s^{1}$ of hydrogen will overlap to the 2p^{3}-orbital of carbon.

Thus, the hybridization of Hydrogen is $s$ -hybridization and the hybridization of Carbon is $sp^{3}$ -hybridization.