Are the properties of Rubidium more similar to those of cesium or those of strontium?

Chemistry

1 answer:

fenix001 [56]3 years ago

5 0

More similar to Cesium

Explanation:

The properties of Rubidium are more similar to those of cesium compared to strontium.

Elements in the same group on the periodic table have similar chemical properties.

Rubidium and Cesium are located in the first group on the periodic table.
Other elements in this group are lithium, sodium, potassium and francium
Strontium belongs to the second group on the periodic table.
The first group have a ns¹ valence shells electronic configuration.
They are all referred to as alkali metals

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View Available Hint(s) Check all that apply. Hydrogen bonding occurs when a hydrogen atom is covalently bonded to an NN, OO, or

Sauron [17]

Answer:

Hydrogen bonding occurs when a hydrogen atom is covalently bonded to an NN, OO, or FF atom.

A hydrogen atom acquires a partial positive charge when it is covalently bonded to an FF atom.

A hydrogen bond is possible with only certain hydrogen-containing compounds.

Explanation:

A hydrogen bond does not occur in all hydrogen containing compounds. Hydrogen bonds only occur in those compounds where hydrogen is bonded to a highly electronegative element such as fluorine, oxygen or nitrogen.

In a hydrogen bonded specie, hydrogen acquires a partial positive charge and the electronegative element acquires a partial negative charge which extends throughout the molecule.

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SeF6 1. Lewis Structure 2. Perspective drawing 3. Number of atoms bonded to central atom 4. Number of non-bonding electron pairs

Anarel [89]

The answer is- $SF_{6}$ is octahedral in electronic and molecular geometry with 6 Fluorine atoms bonded to central atom S.

Lewis structures are the diagrams in which the valence electrons of the atoms of a compound are arranged around the atoms showing the bonding between the atom and the lone pair of electrons existing in the molecule.

Determine the molecular geometry of $SF_{6}$ .

Valence Shell Electron Pair Repulsion theory is commonly known as VSEPR theory and it helps to predict the geometry of molecules.
According to this theory, electrons are arranged around the central atom of the molecule in such a way that there is minimum electrostatic repulsion between these electrons.
Now, calculate the total number of valence electrons in $SF_{6}$ .

$Valence\ electrons\ in\ SF_{6}= Valence\ electrons\ in\ S +\ 6(Valence\ electrons\ in\ F)$

Valence electrons of S = 6

Valence electrons of F = 7

Thus, the valence electrons in $SF_{6}$ are-

$Valence\ number\ of\ electrons\ in\ SF_{6} = (6) + 6(7) = 48\ electrons.$

The Lewis structure of $SF_{6}$ is - (Image attached).
In the structure, the number of atoms bonded to central atom (S) = 6.
Number of non-bonding electron pairs on the central atom = 0 (as all the valence electrons are bonded to F).
Electronic geometry in case of 6 bond pairs is octahedral.
Molecular geometry us also octahedral with bond angles 90°.
Central atom is sp3d2 hybridised.
$SF_{6}$ is a non-polar molecule.

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